Physics Quiz: Calculate Gravitational Force Mathematically

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Calculate Gravitational Force MathematicallyQuestion 1 of 20

A $60\ \text{kg}$ astronaut is on Earth's surface. Using $M_E = 6.0 \times 10^{24}\ \text{kg}$ , $R_E = 6.4 \times 10^{6}\ \text{m}$ , and $G = 6.67 \times 10^{-11}\ \text{N}\cdot\text{m}^2/\text{kg}^2$ , what is the gravitational force on the astronaut according to $F = G\dfrac{M_E m}{R_E^2}$ ?

5.9 \times 10^{2}\ \text{N}

5.9 \times 10^{1}\ \text{N}

9.8 \times 10^{0}\ \text{N}

3.8 \times 10^{9}\ \text{N}

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Physics Quiz: Calculate Gravitational Force Mathematically

Practice Calculate Gravitational Force Mathematically in Physics with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

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This quiz focuses on Calculate Gravitational Force Mathematically, giving you a quick way to practice the rules, question types, and explanations that matter most for Physics.

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Question 1

$5.9 \times 10^{2}\ \text{N}$ (correct answer)
$5.9 \times 10^{1}\ \text{N}$
$9.8 \times 10^{0}\ \text{N}$
$3.8 \times 10^{9}\ \text{N}$

Explanation: This question tests understanding of Newton's Law of Universal Gravitation and the ability to calculate gravitational force between two masses. Newton's Law of Universal Gravitation states that every mass attracts every other mass with a force proportional to the product of their masses and inversely proportional to the square of the distance between their centers: F = G(m₁m₂)/r², where G = 6.67 × 10⁻¹¹ N·m²/kg² is the universal gravitational constant, m₁ and m₂ are the masses in kilograms, r is the center-to-center distance in meters, and F is the attractive force in Newtons. For a person with mass m = 60 kg on Earth's surface, the distance from Earth's center is r = R_E = 6.4 × 10⁶ m. Using F = G(M_E m)/R_E² = (6.67 × 10⁻¹¹)(6.0 × 10²⁴)(60)/(6.4 × 10⁶)² = (6.67 × 10⁻¹¹)(3.6 × 10²⁶)/(4.096 × 10¹³) = (2.401 × 10¹⁶)/(4.096 × 10¹³) = 5.86 × 10² N, which matches the person's weight W = mg = (60)(9.8) ≈ 588 N, confirming that gravity follows the universal law. Choice A is correct because it properly applies F = G(m₁m₂)/r² with correct values and units, yielding approximately 5.9 × 10² N when considering typical rounding. Choice B uses r in the denominator instead of r², calculating F = GMm/r, which misses the inverse square relationship and makes the force too large by a factor of r = 6.4 × 10⁶ m. When calculating gravitational force: (1) identify both masses in kg and distance between centers in m, (2) for objects on a planet's surface use r = radius, for objects in orbit use r = radius + altitude, (3) substitute carefully into F = G(m₁m₂)/r² keeping track of scientific notation, (4) check that the result makes sense: gravitational forces between everyday objects are tiny (~10⁻⁸ N), between person and Earth are hundreds of Newtons (matching weight), and between astronomical objects are enormous (~10²⁰ N or larger). Remember that gravitational force is mutual (Newton's Third Law): Earth pulls on you with the same magnitude force that you pull on Earth, regardless of the huge mass difference—the forces are equal but the accelerations are vastly different because a = F/m, so your acceleration toward Earth (9.8 m/s²) is enormous while Earth's acceleration toward you is negligible due to Earth's massive mass.

Question 2

In the Earth–Moon system, assume the Earth has mass $M_E = 6.0 \times 10^{24}\ \text{kg}$ , the Moon has mass $M_{\text{Moon}} = 7.3 \times 10^{22}\ \text{kg}$ , and the distance between their centers is $r = 3.8 \times 10^{8}\ \text{m}$ . Using Newton's Law of Universal Gravitation, $F = G\dfrac{m_1 m_2}{r^2}$ with $G = 6.67 \times 10^{-11}\ \text{N}\cdot\text{m}^2/\text{kg}^2$ , what is the magnitude of the gravitational force between Earth and the Moon?

$2.0 \times 10^{20}\ \text{N}$ (correct answer)
$2.0 \times 10^{2}\ \text{N}$
$2.0 \times 10^{28}\ \text{N}$
$2.0 \times 10^{11}\ \text{N}$

Explanation: This question tests understanding of Newton's Law of Universal Gravitation and the ability to calculate gravitational force between two masses. Newton's Law of Universal Gravitation states that every mass attracts every other mass with a force proportional to the product of their masses and inversely proportional to the square of the distance between their centers: F = G(m₁m₂)/r², where G = 6.67 × 10⁻¹¹ N·m²/kg² is the universal gravitational constant, m₁ and m₂ are the masses in kilograms, r is the center-to-center distance in meters, and F is the attractive force in Newtons. To find the gravitational force between Earth with mass m₁ = 6.0 × 10²⁴ kg and the Moon with mass m₂ = 7.3 × 10²² kg separated by distance r = 3.8 × 10⁸ m, we substitute into F = G(m₁m₂)/r²: F = (6.67 × 10⁻¹¹ N·m²/kg²)(6.0 × 10²⁴)(7.3 × 10²²)/(3.8 × 10⁸)² = (6.67 × 10⁻¹¹)(4.38 × 10⁴⁷)/(1.444 × 10¹⁷) = 2.02 × 10²⁰ N. Choice A is correct because it properly applies F = G(m₁m₂)/r² with correct values and units, accurately handling scientific notation with the correct power of 10. Choice D has the calculation correct but uses the wrong power of 10 in scientific notation (11 instead of 20), likely from an error in adding/subtracting exponents when multiplying/dividing powers of 10. When calculating gravitational force: (1) identify both masses in kg and distance between centers in m, (2) for objects on a planet's surface use r = radius, for objects in orbit use r = radius + altitude, (3) substitute carefully into F = G(m₁m₂)/r² keeping track of scientific notation, (4) check that the result makes sense: gravitational forces between everyday objects are tiny (~10⁻⁸ N), between person and Earth are hundreds of Newtons (matching weight), and between astronomical objects are enormous (~10²⁰ N or larger). The key insight of the inverse square law is that gravitational force decreases rapidly with distance: double the distance and force drops to 1/4, triple the distance and force drops to 1/9—this r² in the denominator makes gravity a short-range force in practical terms, though it technically extends infinitely.

Question 3

A geostationary satellite of mass $1.0 \times 10^3\ \text{kg}$ orbits Earth at an altitude of $3.6 \times 10^7\ \text{m}$ above the surface. Use $M_E = 6.0 \times 10^{24}\ \text{kg}$ , $R_E = 6.4 \times 10^6\ \text{m}$ , and $G = 6.67 \times 10^{-11}\ \text{N}\cdot\text{m}^2/\text{kg}^2$ . What is the magnitude of the gravitational force on the satellite? (Use $r = R_E + \text{altitude}$ .)

$2.2 \times 10^{2}\ \text{N}$ (correct answer)
$1.5 \times 10^{3}\ \text{N}$
$8.7 \times 10^{2}\ \text{N}$
$3.4 \times 10^{-5}\ \text{N}$

Explanation: This question tests understanding of Newton's Law of Universal Gravitation and the ability to calculate gravitational force on a geostationary satellite. Newton's Law of Universal Gravitation states that every mass attracts every other mass with a force proportional to the product of their masses and inversely proportional to the square of the distance between their centers: F = G(m₁m₂)/r², where G = 6.67 × 10⁻¹¹ N·m²/kg² is the universal gravitational constant, m₁ and m₂ are the masses in kilograms, r is the center-to-center distance in meters, and F is the attractive force in Newtons. A satellite at altitude h = 3.6 × 10⁷ m above Earth's surface is at distance r = R_E + h = 6.4 × 10⁶ + 3.6 × 10⁷ = 4.24 × 10⁷ m from Earth's center. The gravitational force is F = G(M_E m)/r² = (6.67 × 10⁻¹¹)(6.0 × 10²⁴)(1.0 × 10³)/(4.24 × 10⁷)² = (6.67 × 10⁻¹¹)(6.0 × 10²⁷)/(1.798 × 10¹⁵) = 4.002 × 10¹⁷/1.798 × 10¹⁵ = 2.23 × 10² ≈ 2.2 × 10² N, which provides the centripetal force keeping the satellite in orbit. Choice A is correct because it properly applies F = G(m₁m₂)/r² with correct values and correctly uses center-to-center distance including altitude for orbital calculation. Choice B (1.5 × 10³ N) likely uses only Earth's radius when it should use radius plus altitude (r = R_E + h), making the distance too small and the calculated force too large—for orbital calculations, r must be distance from planet's center to satellite. When calculating gravitational force on satellites: (1) always use r = planet radius + altitude for the distance, (2) geostationary satellites are much farther than low Earth orbit satellites, so forces are smaller, (3) this gravitational force provides the centripetal acceleration needed for circular orbital motion.

Question 4

Earth and the Moon attract each other gravitationally. Use $M_E = 6.0 \times 10^{24}\ \text{kg}$ , $M_{\text{Moon}} = 7.3 \times 10^{22}\ \text{kg}$ , center-to-center distance $r = 3.8 \times 10^8\ \text{m}$ , and $G = 6.67 \times 10^{-11}\ \text{N}\cdot\text{m}^2/\text{kg}^2$ . What is the magnitude of the gravitational force between Earth and the Moon?

$2.0 \times 10^{20}\ \text{N}$ (correct answer)
$2.0 \times 10^{12}\ \text{N}$
$5.8 \times 10^{2}\ \text{N}$
$7.6 \times 10^{28}\ \text{N}$

Explanation: This question tests understanding of Newton's Law of Universal Gravitation and the ability to calculate gravitational force between astronomical objects. Newton's Law of Universal Gravitation states that every mass attracts every other mass with a force proportional to the product of their masses and inversely proportional to the square of the distance between their centers: F = G(m₁m₂)/r², where G = 6.67 × 10⁻¹¹ N·m²/kg² is the universal gravitational constant, m₁ and m₂ are the masses in kilograms, r is the center-to-center distance in meters, and F is the attractive force in Newtons. To find the gravitational force between Earth with mass M_E = 6.0 × 10²⁴ kg and the Moon with mass M_Moon = 7.3 × 10²² kg separated by distance r = 3.8 × 10⁸ m, we substitute into F = G(m₁m₂)/r²: F = (6.67 × 10⁻¹¹ N·m²/kg²)(6.0 × 10²⁴)(7.3 × 10²²)/(3.8 × 10⁸)² = (6.67 × 10⁻¹¹)(4.38 × 10⁴⁷)/(1.444 × 10¹⁷) = 2.92 × 10³⁷/1.444 × 10¹⁷ = 2.0 × 10²⁰ N. Choice A is correct because it properly applies F = G(m₁m₂)/r² with correct values and accurately handles scientific notation with correct power of 10. Choice B (2.0 × 10¹² N) has the calculation correct but uses the wrong power of 10 in scientific notation (10¹² instead of 10²⁰), likely from an error in adding/subtracting exponents when multiplying/dividing powers of 10. When calculating gravitational force between astronomical objects: (1) identify both masses in kg and distance between centers in m, (2) substitute carefully into F = G(m₁m₂)/r² keeping track of scientific notation, (3) check that the result makes sense: gravitational forces between astronomical objects are enormous (~10²⁰ N or larger). Remember that gravitational force is mutual (Newton's Third Law): Earth pulls on the Moon with the same magnitude force that the Moon pulls on Earth, regardless of the huge mass difference—the forces are equal but the accelerations are vastly different because a = F/m.

Question 5

A probe of mass $5.0 \times 10^3\ \text{kg}$ is near Earth. Use $M_E = 6.0 \times 10^{24}\ \text{kg}$ , $R_E = 6.4 \times 10^6\ \text{m}$ , and $G = 6.67 \times 10^{-11}\ \text{N}\cdot\text{m}^2/\text{kg}^2$ . What is the magnitude of the gravitational force that the probe exerts on Earth when the probe is at Earth's surface (take $r = R_E$ )?

$4.9 \times 10^{4}\ \text{N}$ (correct answer)
$2.9 \times 10^{-5}\ \text{N}$
$3.1 \times 10^{11}\ \text{N}$
$7.8 \times 10^{4}\ \text{N}$

Explanation: This question tests understanding of Newton's Law of Universal Gravitation and Newton's Third Law, specifically asking for the force the probe exerts on Earth. Newton's Law of Universal Gravitation states that every mass attracts every other mass with a force proportional to the product of their masses and inversely proportional to the square of the distance between their centers: F = G(m₁m₂)/r², where G = 6.67 × 10⁻¹¹ N·m²/kg² is the universal gravitational constant, m₁ and m₂ are the masses in kilograms, r is the center-to-center distance in meters, and F is the attractive force in Newtons. For a probe with mass m = 5.0 × 10³ kg on Earth's surface, the distance from Earth's center is r = R_E = 6.4 × 10⁶ m. Using F = G(M_E m)/R_E² = (6.67 × 10⁻¹¹)(6.0 × 10²⁴)(5.0 × 10³)/(6.4 × 10⁶)² = (6.67 × 10⁻¹¹)(3.0 × 10²⁸)/(4.096 × 10¹³) = 2.001 × 10¹⁸/4.096 × 10¹³ = 4.88 × 10⁴ ≈ 4.9 × 10⁴ N. Choice A is correct because by Newton's Third Law, the force Earth exerts on the probe equals the force the probe exerts on Earth—both objects experience equal magnitude forces (though in opposite directions). Choice C (3.1 × 10¹¹ N) incorrectly suggests the more massive object experiences a larger gravitational force, when actually by Newton's Third Law both objects experience equal magnitude forces regardless of their mass difference. Remember that gravitational force is mutual (Newton's Third Law): Earth pulls on the probe with the same magnitude force that the probe pulls on Earth—the forces are equal but the accelerations are vastly different because a = F/m, so the probe's acceleration toward Earth (9.8 m/s²) is enormous while Earth's acceleration toward the probe is negligible due to Earth's massive mass.

Question 6

A spacecraft of mass $m = 2.0 \times 10^3\ \text{kg}$ orbits Earth at an altitude of $4.0 \times 10^5\ \text{m}$ above Earth's surface. Use Newton's Law of Universal Gravitation, $F = \dfrac{G M_E m}{r^2}$ , with $G = 6.67 \times 10^{-11}\ \text{N}\cdot\text{m}^2/\text{kg}^2$ , $M_E = 6.0 \times 10^{24}\ \text{kg}$ , and $R_E = 6.4 \times 10^6\ \text{m}$ . What is the magnitude of the gravitational force on the spacecraft? (Use $r = R_E + \text{altitude}$ .)

$2.8 \times 10^4\ \text{N}$
$1.9 \times 10^3\ \text{N}$
$1.8 \times 10^4\ \text{N}$ (correct answer)
$1.2 \times 10^{11}\ \text{N}$

Explanation: This question tests understanding of Newton's Law of Universal Gravitation and the ability to calculate gravitational force on an orbiting spacecraft. Newton's Law of Universal Gravitation states that every mass attracts every other mass with a force proportional to the product of their masses and inversely proportional to the square of the distance between their centers: F = G(m₁m₂)/r², where G = 6.67 × 10⁻¹¹ N·m²/kg² is the universal gravitational constant, m₁ and m₂ are the masses in kilograms, r is the center-to-center distance in meters, and F is the attractive force in Newtons. For a spacecraft at altitude h = 4.0 × 10⁵ m above Earth's surface, the distance from Earth's center is r = R_E + h = 6.4 × 10⁶ + 4.0 × 10⁵ = 6.8 × 10⁶ m. The gravitational force is F = G(M_E m)/r² = (6.67 × 10⁻¹¹)(6.0 × 10²⁴)(2.0 × 10³)/(6.8 × 10⁶)² = (6.67 × 10⁻¹¹)(1.2 × 10²⁸)/(4.624 × 10¹³) = 8.004 × 10¹⁷/4.624 × 10¹³ = 1.73 × 10⁴ ≈ 1.8 × 10⁴ N. Choice C is correct because it properly applies F = G(M_E m)/r² with the correct center-to-center distance r = R_E + altitude = 6.8 × 10⁶ m and accurately handles the scientific notation. Choice A (2.8 × 10⁴ N) likely uses an incorrect distance calculation, while Choice B (1.9 × 10³ N) has the wrong power of 10, and Choice D (1.2 × 10¹¹ N) is far too large, possibly from omitting the r² term in the denominator. When calculating gravitational force on satellites: (1) always use r = planet radius + altitude for the center-to-center distance, (2) substitute carefully into F = G(M_planet m_satellite)/r² keeping track of scientific notation, (3) verify the result is reasonable—satellites in low Earth orbit experience forces in the 10³-10⁴ N range. The key insight is that even at 400 km altitude, the gravitational force is still substantial, providing the centripetal force that keeps the spacecraft in orbit rather than flying off into space.

Question 7

A student wants to verify that weight near Earth's surface can be found using Newton's Law of Universal Gravitation. Use $m = 70\ \text{kg}$ , $M_E = 6.0 \times 10^{24}\ \text{kg}$ , $R_E = 6.4 \times 10^6\ \text{m}$ , and $G = 6.67 \times 10^{-11}\ \text{N}\cdot\text{m}^2/\text{kg}^2$ . What is the magnitude of the gravitational force on the student at Earth's surface (take $r = R_E$ )?

$6.9 \times 10^{2}\ \text{N}$ (correct answer)
$1.1 \times 10^{-7}\ \text{N}$
$4.4 \times 10^{9}\ \text{N}$
$1.1 \times 10^{3}\ \text{N}$

Explanation: This question tests understanding of Newton's Law of Universal Gravitation and verifies that weight near Earth's surface can be calculated using the universal gravitational formula. Newton's Law of Universal Gravitation states that every mass attracts every other mass with a force proportional to the product of their masses and inversely proportional to the square of the distance between their centers: F = G(m₁m₂)/r², where G = 6.67 × 10⁻¹¹ N·m²/kg² is the universal gravitational constant, m₁ and m₂ are the masses in kilograms, r is the center-to-center distance in meters, and F is the attractive force in Newtons. For a person with mass m = 70 kg on Earth's surface, the distance from Earth's center is r = R_E = 6.4 × 10⁶ m. Using F = G(M_E m)/R_E² = (6.67 × 10⁻¹¹)(6.0 × 10²⁴)(70)/(6.4 × 10⁶)² = (6.67 × 10⁻¹¹)(4.2 × 10²⁶)/(4.096 × 10¹³) = 2.8 × 10¹⁶/4.096 × 10¹³ = 6.84 × 10² ≈ 6.9 × 10² N, which matches the person's weight W = mg = (70)(9.8) ≈ 686 N, confirming that gravity follows the universal law. Choice A is correct because it properly applies F = G(m₁m₂)/r² with correct values and verifies that calculated gravitational force matches weight W = mg. Choice B (1.1 × 10⁻⁷ N) likely forgets to include the gravitational constant G or uses an incorrect value, leading to a result that's off by several orders of magnitude—a 70 kg person should weigh hundreds of Newtons, not a tiny fraction of a Newton. When calculating gravitational force at a planet's surface: (1) use r = planet's radius for the distance, (2) the calculated force should equal the object's weight W = mg, (3) this verification shows that g = GM/R² at the surface. This connection between Newton's universal law and everyday weight demonstrates the universality of gravitational physics.

Question 8

A person of mass $m = 70\ \text{kg}$ is standing on Earth's surface. Use $M_E = 6.0 \times 10^{24}\ \text{kg}$ , $R_E = 6.4 \times 10^6\ \text{m}$ , and $G = 6.67 \times 10^{-11}\ \text{N}\cdot\text{m}^2/\text{kg}^2$ . Treat Earth as a sphere and take $r = R_E$ . Using $F = G\dfrac{M_E m}{r^2}$ , what is the magnitude of the gravitational force on the person (their weight)?

$6.9 \times 10^2\ \text{N}$ (correct answer)
$6.9 \times 10^1\ \text{N}$
$6.9 \times 10^3\ \text{N}$
$1.1 \times 10^{-8}\ \text{N}$

Explanation: This question tests understanding of Newton's Law of Universal Gravitation and the ability to calculate gravitational force between two masses. Newton's Law of Universal Gravitation states that every mass attracts every other mass with a force proportional to the product of their masses and inversely proportional to the square of the distance between their centers: F = G(m₁m₂)/r², where G = 6.67 × 10⁻¹¹ N·m²/kg² is the universal gravitational constant, m₁ and m₂ are the masses in kilograms, r is the center-to-center distance in meters, and F is the attractive force in Newtons. For a person with mass m = 70 kg on Earth's surface, the distance from Earth's center is r = R_E = 6.4 × 10⁶ m. Using F = G(M_E m)/R_E² = (6.67 × 10⁻¹¹)(6.0 × 10²⁴)(70)/(6.4 × 10⁶)² = (6.67 × 10⁻¹¹)(4.2 × 10²⁶)/(4.096 × 10¹³) = (2.801 × 10¹⁶)/(4.096 × 10¹³) ≈ 6.84 × 10² N, which matches the person's weight W = mg = (70)(9.8) ≈ 686 N, confirming that gravity follows the universal law. Choice A is correct because it properly applies F = G(m₁m₂)/r² with correct values and units, verifying that calculated gravitational force matches weight W = mg. Choice D is a tempting distractor but fails because it forgets to include the gravitational constant G or uses an incorrect value, leading to a result that's off by several orders of magnitude. When calculating gravitational force: (1) identify both masses in kg and distance between centers in m, (2) for objects on a planet's surface use r = radius, (3) substitute carefully into F = G(m₁m₂)/r² keeping track of scientific notation, (4) check that the result makes sense: between person and Earth are hundreds of Newtons (matching weight). Remember that gravitational force is mutual (Newton's Third Law): Earth pulls on you with the same magnitude force that you pull on Earth, regardless of the huge mass difference—the forces are equal but the accelerations are vastly different because a = F/m, so your acceleration toward Earth (9.8 m/s²) is enormous while Earth's acceleration toward you is negligible due to Earth's massive mass.

Question 9

A student calculates the gravitational force between Earth ( $M_E = 6.0 \times 10^{24}\ \text{kg}$ ) and a $60\ \text{kg}$ person at Earth's surface ( $r = R_E = 6.4 \times 10^6\ \text{m}$ ) using $G = 6.67 \times 10^{-11}\ \text{N}\cdot\text{m}^2/\text{kg}^2$ . Which value is closest to the correct force magnitude?

$5.9 \times 10^1\ \text{N}$
$5.9 \times 10^2\ \text{N}$ (correct answer)
$5.9 \times 10^3\ \text{N}$
$5.9 \times 10^{-8}\ \text{N}$

Explanation: This question tests understanding of Newton's Law of Universal Gravitation and the ability to calculate gravitational force between two masses. Newton's Law of Universal Gravitation states that every mass attracts every other mass with a force proportional to the product of their masses and inversely proportional to the square of the distance between their centers: F = G(m₁m₂)/r², where G = 6.67 × 10⁻¹¹ N·m²/kg² is the universal gravitational constant, m₁ and m₂ are the masses in kilograms, r is the center-to-center distance in meters, and F is the attractive force in Newtons. For a person with mass m = 60 kg on Earth's surface, the distance from Earth's center is r = R_E = 6.4 × 10⁶ m. Using F = G(M_E m)/R_E² = (6.67 × 10⁻¹¹)(6.0 × 10²⁴)(60)/(6.4 × 10⁶)² = (6.67 × 10⁻¹¹)(3.6 × 10²⁶)/(4.096 × 10¹³) = (2.401 × 10¹⁶)/(4.096 × 10¹³) ≈ 5.86 × 10² N, which matches the person's weight W = mg = (60)(9.8) ≈ 588 N, confirming that gravity follows the universal law. Choice B is correct because it properly applies F = G(m₁m₂)/r² with correct values and units, verifying that calculated gravitational force matches weight W = mg approximated to 5.9 × 10² N. Choice D is a tempting distractor but fails because it forgets to include the gravitational constant G or uses an incorrect value, leading to a result that's off by several orders of magnitude. When calculating gravitational force: (1) identify both masses in kg and distance between centers in m, (2) for objects on a planet's surface use r = radius, (3) substitute carefully into F = G(m₁m₂)/r² keeping track of scientific notation, (4) check that the result makes sense: between person and Earth are hundreds of Newtons (matching weight). Remember that gravitational force is mutual (Newton's Third Law): Earth pulls on you with the same magnitude force that you pull on Earth, regardless of the huge mass difference—the forces are equal but the accelerations are vastly different because a = F/m, so your acceleration toward Earth (9.8 m/s²) is enormous while Earth's acceleration toward you is negligible due to Earth's massive mass.

Question 10

In the Earth–Moon system, the mass of Earth is $M_E = 6.0 \times 10^{24}\ \text{kg}$ and the mass of the Moon is $M_{\text{Moon}} = 7.3 \times 10^{22}\ \text{kg}$ . The distance between their centers is $r = 3.8 \times 10^{8}\ \text{m}$ . Using Newton's Law of Universal Gravitation, $F = G\dfrac{m_1 m_2}{r^2}$ with $G = 6.67 \times 10^{-11}\ \text{N}\cdot\text{m}^2/\text{kg}^2$ , what is the magnitude of the gravitational force between Earth and the Moon?

$2.0 \times 10^{11}\ \text{N}$
$2.0 \times 10^{20}\ \text{N}$ (correct answer)
$7.6 \times 10^{28}\ \text{N}$
$5.0 \times 10^{19}\ \text{N}$

Explanation: This question tests understanding of Newton's Law of Universal Gravitation and the ability to calculate gravitational force between two masses. Newton's Law of Universal Gravitation states that every mass attracts every other mass with a force proportional to the product of their masses and inversely proportional to the square of the distance between their centers: F = G(m₁m₂)/r², where G = 6.67 × 10⁻¹¹ N·m²/kg² is the universal gravitational constant, m₁ and m₂ are the masses in kilograms, r is the center-to-center distance in meters, and F is the attractive force in Newtons. To find the gravitational force between Earth with mass m₁ = 6.0 × 10²⁴ kg and the Moon with mass m₂ = 7.3 × 10²² kg separated by distance r = 3.8 × 10⁸ m, we substitute into F = G(m₁m₂)/r²: F = (6.67 × 10⁻¹¹ N·m²/kg²)(6.0 × 10²⁴ kg)(7.3 × 10²² kg)/(3.8 × 10⁸ m)² = (6.67 × 10⁻¹¹)(4.38 × 10⁴⁷)/(1.444 × 10¹⁷) = (2.92 × 10³⁷)/(1.444 × 10¹⁷) = 2.02 × 10²⁰ N. Choice B is correct because it properly applies F = G(m₁m₂)/r² with correct values and units, handling scientific notation accurately to yield 2.0 × 10²⁰ N. Choice A uses an incorrect power of 10 in scientific notation (11 instead of 20), likely from an error in subtracting exponents when dividing powers of 10. When calculating gravitational force: (1) identify both masses in kg and distance between centers in m, (2) for objects on a planet's surface use r = radius, for objects in orbit use r = radius + altitude, (3) substitute carefully into F = G(m₁m₂)/r² keeping track of scientific notation, (4) check that the result makes sense: gravitational forces between everyday objects are tiny (~10⁻⁸ N), between person and Earth are hundreds of Newtons (matching weight), and between astronomical objects are enormous (~10²⁰ N or larger). The key insight of the inverse square law is that gravitational force decreases rapidly with distance: double the distance and force drops to 1/4, triple the distance and force drops to 1/9—this r² in the denominator makes gravity a short-range force in practical terms, though it technically extends infinitely.

Question 11

Two identical research probes, each of mass $2.0 \times 10^{3}\ \text{kg}$ , are drifting in deep space. Their centers are separated by $r = 10\ \text{m}$ . Using $G = 6.67 \times 10^{-11}\ \text{N}\cdot\text{m}^2/\text{kg}^2$ and $F = G\dfrac{m_1 m_2}{r^2}$ , what is the magnitude of the gravitational force between them?

$2.7 \times 10^{-6}\ \text{N}$ (correct answer)
$2.7 \times 10^{-3}\ \text{N}$
$2.7 \times 10^{-9}\ \text{N}$
$2.7 \times 10^{-7}\ \text{N}$

Explanation: This question tests understanding of Newton's Law of Universal Gravitation and the ability to calculate gravitational force between two masses. Newton's Law of Universal Gravitation states that every mass attracts every other mass with a force proportional to the product of their masses and inversely proportional to the square of the distance between their centers: F = G(m₁m₂)/r², where G = 6.67 × 10⁻¹¹ N·m²/kg² is the universal gravitational constant, m₁ and m₂ are the masses in kilograms, r is the center-to-center distance in meters, and F is the attractive force in Newtons. To find the gravitational force between two probes each with mass m = 2.0 × 10³ kg separated by distance r = 10 m, we substitute into F = G(m₁m₂)/r²: F = (6.67 × 10⁻¹¹ N·m²/kg²)(2.0 × 10³ kg)(2.0 × 10³ kg)/(10 m)² = (6.67 × 10⁻¹¹)(4.0 × 10⁶)/100 = (2.668 × 10⁻⁴)/100 = 2.668 × 10⁻⁶ N. Choice A is correct because it properly applies F = G(m₁m₂)/r² with correct values and units, handling scientific notation accurately to yield 2.7 × 10⁻⁶ N. Choice C has the calculation correct but uses the wrong power of 10 in scientific notation (-9 instead of -6), likely from an error in adding/subtracting exponents when multiplying/dividing powers of 10. When calculating gravitational force: (1) identify both masses in kg and distance between centers in m, (2) for objects on a planet's surface use r = radius, for objects in orbit use r = radius + altitude, (3) substitute carefully into F = G(m₁m₂)/r² keeping track of scientific notation, (4) check that the result makes sense: gravitational forces between everyday objects are tiny (~10⁻⁸ N), between person and Earth are hundreds of Newtons (matching weight), and between astronomical objects are enormous (~10²⁰ N or larger). Remember that gravitational force is mutual (Newton's Third Law): Earth pulls on you with the same magnitude force that you pull on Earth, regardless of the huge mass difference—the forces are equal but the accelerations are vastly different because a = F/m, so your acceleration toward Earth (9.8 m/s²) is enormous while Earth's acceleration toward you is negligible due to Earth's massive mass.

Question 12

A $1.0 \times 10^{3}\ \text{kg}$ spacecraft is $r = 1.0 \times 10^{7}\ \text{m}$ from the center of Earth. Use $M_E = 6.0 \times 10^{24}\ \text{kg}$ and $G = 6.67 \times 10^{-11}\ \text{N}\cdot\text{m}^2/\text{kg}^2$ . What is the magnitude of the gravitational force Earth exerts on the spacecraft (and, by Newton's Third Law, the magnitude the spacecraft exerts on Earth)?

$4.0 \times 10^{3}\ \text{N}$ (correct answer)
$4.0 \times 10^{6}\ \text{N}$
$4.0 \times 10^{2}\ \text{N}$
$4.0 \times 10^{-8}\ \text{N}$

Explanation: This question tests understanding of Newton's Law of Universal Gravitation and the ability to calculate gravitational force between two masses. Newton's Law of Universal Gravitation states that every mass attracts every other mass with a force proportional to the product of their masses and inversely proportional to the square of the distance between their centers: F = G(m₁m₂)/r², where G = 6.67 × 10⁻¹¹ N·m²/kg² is the universal gravitational constant, m₁ and m₂ are the masses in kilograms, r is the center-to-center distance in meters, and F is the attractive force in Newtons. For satellite in orbit: A satellite at altitude h but here directly given r = 1.0 × 10⁷ m from Earth's center. The gravitational force is F = G(M_E m)/r² = (6.67 × 10⁻¹¹)(6.0 × 10²⁴)(1.0 × 10³)/(1.0 × 10⁷)² = (6.67 × 10⁻¹¹)(6.0 × 10²⁷)/(1.0 × 10¹⁴) = (4.002 × 10¹⁷)/(1.0 × 10¹⁴) = 4.002 × 10³ N, which provides the centripetal force keeping the satellite in orbit. Choice A is correct because it properly applies F = G(m₁m₂)/r² with correct values and units, handling scientific notation accurately to yield 4.0 × 10³ N. Choice D has the calculation correct but uses the wrong power of 10 in scientific notation (-8 instead of 3), likely from an error in adding/subtracting exponents when multiplying/dividing powers of 10. When calculating gravitational force: (1) identify both masses in kg and distance between centers in m, (2) for objects on a planet's surface use r = radius, for objects in orbit use r = radius + altitude, (3) substitute carefully into F = G(m₁m₂)/r² keeping track of scientific notation, (4) check that the result makes sense: gravitational forces between everyday objects are tiny (~10⁻⁸ N), between person and Earth are hundreds of Newtons (matching weight), and between astronomical objects are enormous (~10²⁰ N or larger). The key insight of the inverse square law is that gravitational force decreases rapidly with distance: double the distance and force drops to 1/4, triple the distance and force drops to 1/9—this r² in the denominator makes gravity a short-range force in practical terms, though it technically extends infinitely.

Question 13

A $500\ \text{kg}$ satellite is moved from a circular orbit of altitude $400\ \text{km}$ to a circular orbit of altitude $800\ \text{km}$ above Earth's surface. Use $R_E = 6.4 \times 10^{6}\ \text{m}$ . What is the ratio of the gravitational force in the higher orbit to the gravitational force in the lower orbit? (Use $F \propto 1/r^2$ with $r = R_E + h$ .)

$\left(\dfrac{6.8 \times 10^{6}}{7.2 \times 10^{6}}\right)^2 \approx 0.89$ (correct answer)
$\dfrac{6.8 \times 10^{6}}{7.2 \times 10^{6}} \approx 0.94$
$\left(\dfrac{7.2 \times 10^{6}}{6.8 \times 10^{6}}\right)^2 \approx 1.12$
$\left(\dfrac{6.4 \times 10^{6}}{8.0 \times 10^{5}}\right)^2 \approx 64$

Explanation: This question tests understanding of Newton's Law of Universal Gravitation and the ability to calculate gravitational force between two masses. Newton's Law of Universal Gravitation states that every mass attracts every other mass with a force proportional to the product of their masses and inversely proportional to the square of the distance between their centers: F = G(m₁m₂)/r², where G = 6.67 × 10⁻¹¹ N·m²/kg² is the universal gravitational constant, m₁ and m₂ are the masses in kilograms, r is the center-to-center distance in meters, and F is the attractive force in Newtons. For distance comparison: If the distance between two masses changes from r to 2r, the force changes by a factor of (r/2r)² = 1/4, so the new force is F_new = F_original × (r₁/r₂)² = [original] × (1/2)² = [original] × 1/4 = [result], demonstrating the inverse square relationship. Choice A is correct because it properly applies F = G(m₁m₂)/r² with correct values and units / correctly uses center-to-center distance including altitude for orbital calculation / accurately applies inverse square law showing force ratio / properly handles scientific notation with correct power of 10 / verifies that calculated gravitational force matches weight W = mg. Choice C incorrectly claims the force [doubles/halves] when the distance [doubles/halves], missing the inverse square relationship—when distance doubles, force becomes 1/4 (not 1/2), and when distance halves, force becomes 4 times larger (not 2 times). When calculating gravitational force: (1) identify both masses in kg and distance between centers in m, (2) for objects on a planet's surface use r = radius, for objects in orbit use r = radius + altitude, (3) substitute carefully into F = G(m₁m₂)/r² keeping track of scientific notation, (4) check that the result makes sense: gravitational forces between everyday objects are tiny (~10⁻⁸ N), between person and Earth are hundreds of Newtons (matching weight), and between astronomical objects are enormous (~10²⁰ N or larger). The key insight of the inverse square law is that gravitational force decreases rapidly with distance: double the distance and force drops to 1/4, triple the distance and force drops to 1/9—this r² in the denominator makes gravity a short-range force in practical terms, though it technically extends infinitely.

Question 14

A $70\ \text{kg}$ person is standing on Earth's surface. Use $M_E = 6.0 \times 10^{24}\ \text{kg}$ , $R_E = 6.4 \times 10^{6}\ \text{m}$ , and $G = 6.67 \times 10^{-11}\ \text{N}\cdot\text{m}^2/\text{kg}^2$ . Using $F = G\dfrac{M_E m}{R_E^2}$ , what is the magnitude of the gravitational force (weight) on the person?

$6.9 \times 10^{2}\ \text{N}$ (correct answer)
$1.1 \times 10^{2}\ \text{N}$
$4.4 \times 10^{9}\ \text{N}$
$6.9 \times 10^{1}\ \text{N}$

Explanation: This question tests understanding of Newton's Law of Universal Gravitation and the ability to calculate gravitational force between two masses. Newton's Law of Universal Gravitation states that every mass attracts every other mass with a force proportional to the product of their masses and inversely proportional to the square of the distance between their centers: F = G(m₁m₂)/r², where G = 6.67 × 10⁻¹¹ N·m²/kg² is the universal gravitational constant, m₁ and m₂ are the masses in kilograms, r is the center-to-center distance in meters, and F is the attractive force in Newtons. For a person with mass m = 70 kg on Earth's surface, the distance from Earth's center is r = R_E = 6.4 × 10⁶ m. Using F = G(M_E m)/R_E² = (6.67 × 10⁻¹¹)(6.0 × 10²⁴)(70)/(6.4 × 10⁶)² = (6.67 × 10⁻¹¹)(4.2 × 10²⁶)/(4.096 × 10¹³) = (2.801 × 10¹⁶)/(4.096 × 10¹³) = 6.84 × 10² N, which matches the person's weight W = mg = (70)(9.8) ≈ 686 N, confirming that gravity follows the universal law. Choice A is correct because it properly applies F = G(m₁m₂)/r² with correct values and units, yielding approximately 6.9 × 10² N when considering typical rounding or slight variations in constants. Choice D forgets to include the gravitational constant G or uses an incorrect value, leading to a result that's off by several orders of magnitude. When calculating gravitational force: (1) identify both masses in kg and distance between centers in m, (2) for objects on a planet's surface use r = radius, for objects in orbit use r = radius + altitude, (3) substitute carefully into F = G(m₁m₂)/r² keeping track of scientific notation, (4) check that the result makes sense: gravitational forces between everyday objects are tiny (~10⁻⁸ N), between person and Earth are hundreds of Newtons (matching weight), and between astronomical objects are enormous (~10²⁰ N or larger). Remember that gravitational force is mutual (Newton's Third Law): Earth pulls on you with the same magnitude force that you pull on Earth, regardless of the huge mass difference—the forces are equal but the accelerations are vastly different because a = F/m, so your acceleration toward Earth (9.8 m/s²) is enormous while Earth's acceleration toward you is negligible due to Earth's massive mass.

Question 15

Two spacecraft with masses $m_1 = 1.0 \times 10^{4}\ \text{kg}$ and $m_2 = 2.0 \times 10^{4}\ \text{kg}$ are separated by $r = 100\ \text{m}$ (center to center). If the gravitational force magnitude between them is $F$ , what is the new force magnitude when the separation increases to $200\ \text{m}$ ? (Use the inverse-square relationship in $F = G\dfrac{m_1 m_2}{r^2}$ .)

$4F$
$2F$
$\dfrac{1}{2}F$
$\dfrac{1}{4}F$ (correct answer)

Explanation: This question tests understanding of Newton's Law of Universal Gravitation and the ability to calculate gravitational force between two masses. Newton's Law of Universal Gravitation states that every mass attracts every other mass with a force proportional to the product of their masses and inversely proportional to the square of the distance between their centers: F = G(m₁m₂)/r², where G = 6.67 × 10⁻¹¹ N·m²/kg² is the universal gravitational constant, m₁ and m₂ are the masses in kilograms, r is the center-to-center distance in meters, and F is the attractive force in Newtons. For distance comparison: If the distance between two masses changes from r to 2r, the force changes by a factor of (r/2r)² = 1/4, so the new force is F_new = F_original × (r₁/r₂)² = F × (1/2)² = F × 1/4 = (1/4)F, demonstrating the inverse square relationship. Choice D is correct because it accurately applies inverse square law showing force ratio. Choice C incorrectly claims the force halves when the distance doubles, missing the inverse square relationship—when distance doubles, force becomes 1/4 (not 1/2), and when distance halves, force becomes 4 times larger (not 2 times). When calculating gravitational force: (1) identify both masses in kg and distance between centers in m, (2) for objects on a planet's surface use r = radius, for objects in orbit use r = radius + altitude, (3) substitute carefully into F = G(m₁m₂)/r² keeping track of scientific notation, (4) check that the result makes sense: gravitational forces between everyday objects are tiny (~10⁻⁸ N), between person and Earth are hundreds of Newtons (matching weight), and between astronomical objects are enormous (~10²⁰ N or larger). The key insight of the inverse square law is that gravitational force decreases rapidly with distance: double the distance and force drops to 1/4, triple the distance and force drops to 1/9—this r² in the denominator makes gravity a short-range force in practical terms, though it technically extends infinitely.

Question 16

Two asteroids of masses $m_1 = 3.0 \times 10^{12}\ \text{kg}$ and $m_2 = 6.0 \times 10^{12}\ \text{kg}$ are separated by $r = 2.0 \times 10^{6}\ \text{m}$ (center to center). Using $G = 6.67 \times 10^{-11}\ \text{N}\cdot\text{m}^2/\text{kg}^2$ , what is the magnitude of the gravitational force between them?

$3.0 \times 10^{8}\ \text{N}$
$3.0 \times 10^{2}\ \text{N}$ (correct answer)
$3.0 \times 10^{-4}\ \text{N}$
$1.5 \times 10^{11}\ \text{N}$

Explanation: This question tests understanding of Newton's Law of Universal Gravitation and the ability to calculate gravitational force between two masses. Newton's Law of Universal Gravitation states that every mass attracts every other mass with a force proportional to the product of their masses and inversely proportional to the square of the distance between their centers: F = G(m₁m₂)/r², where G = 6.67 × 10⁻¹¹ N·m²/kg² is the universal gravitational constant, m₁ and m₂ are the masses in kilograms, r is the center-to-center distance in meters, and F is the attractive force in Newtons. To find the gravitational force between asteroid 1 with mass m₁ = 3.0 × 10¹² kg and asteroid 2 with mass m₂ = 6.0 × 10¹² kg separated by distance r = 2.0 × 10⁶ m, we substitute into F = G(m₁m₂)/r²: F = (6.67 × 10⁻¹¹ N·m²/kg²)(1.8 × 10²⁵)/(2.0 × 10⁶)² = (6.67 × 10⁻¹¹)(1.8 × 10²⁵)/(4.0 × 10¹²) = 3.0 × 10² N. Choice B is correct because it properly applies F = G(m₁m₂)/r² with correct values and units, handling scientific notation accurately with the correct power of 10. Choice A has the calculation correct but uses the wrong power of 10 in scientific notation (8 instead of 2), likely from an error in adding/subtracting exponents when multiplying/dividing powers of 10. When calculating gravitational force: (1) identify both masses in kg and distance between centers in m, (2) for objects on a planet's surface use r = radius, for objects in orbit use r = radius + altitude, (3) substitute carefully into F = G(m₁m₂)/r² keeping track of scientific notation, (4) check that the result makes sense: gravitational forces between everyday objects are tiny (~10⁻⁸ N), between person and Earth are hundreds of Newtons (matching weight), and between astronomical objects are enormous (~10²⁰ N or larger). The key insight of the inverse square law is that gravitational force decreases rapidly with distance: double the distance and force drops to 1/4, triple the distance and force drops to 1/9—this r² in the denominator makes gravity a short-range force in practical terms, though it technically extends infinitely.

Question 17

A geostationary satellite of mass $m = 2.0 \times 10^{3}\ \text{kg}$ orbits Earth at an orbital radius (center to center) of $r = 4.2 \times 10^{7}\ \text{m}$ . Use $M_E = 6.0 \times 10^{24}\ \text{kg}$ and $G = 6.67 \times 10^{-11}\ \text{N}\cdot\text{m}^2/\text{kg}^2$ . What is the magnitude of the gravitational force on the satellite?

$4.5 \times 10^{2}\ \text{N}$ (correct answer)
$1.9 \times 10^{4}\ \text{N}$
$1.1 \times 10^{-7}\ \text{N}$
$1.9 \times 10^{11}\ \text{N}$

Explanation: This question tests understanding of Newton's Law of Universal Gravitation and the ability to calculate gravitational force between two masses. Newton's Law of Universal Gravitation states that every mass attracts every other mass with a force proportional to the product of their masses and inversely proportional to the square of the distance between their centers: F = G(m₁m₂)/r², where G = 6.67 × 10⁻¹¹ N·m²/kg² is the universal gravitational constant, m₁ and m₂ are the masses in kilograms, r is the center-to-center distance in meters, and F is the attractive force in Newtons. A satellite at altitude h (but given as r = 4.2 × 10⁷ m from center) uses F = G(M_E m)/r² = (6.67 × 10⁻¹¹)(6.0 × 10²⁴)(2.0 × 10³)/(4.2 × 10⁷)² = (6.67 × 10⁻¹¹)(1.2 × 10²⁸)/(1.764 × 10¹⁵) = 4.5 × 10² N, which provides the centripetal force keeping the satellite in orbit. Choice A is correct because it correctly uses center-to-center distance including altitude for orbital calculation. Choice B uses only Earth's radius when it should use radius plus altitude (r = R_E + h), making the distance too small and the calculated force too large—for orbital calculations, r must be distance from planet's center to satellite. When calculating gravitational force: (1) identify both masses in kg and distance between centers in m, (2) for objects on a planet's surface use r = radius, for objects in orbit use r = radius + altitude, (3) substitute carefully into F = G(m₁m₂)/r² keeping track of scientific notation, (4) check that the result makes sense: gravitational forces between everyday objects are tiny (~10⁻⁸ N), between person and Earth are hundreds of Newtons (matching weight), and between astronomical objects are enormous (~10²⁰ N or larger). The key insight of the inverse square law is that gravitational force decreases rapidly with distance: double the distance and force drops to 1/4, triple the distance and force drops to 1/9—this r² in the denominator makes gravity a short-range force in practical terms, though it technically extends infinitely.

Question 18

In the Earth–Moon system, take $M_E = 6.0 \times 10^{24}\ \text{kg}$ , $M_{\text{Moon}} = 7.3 \times 10^{22}\ \text{kg}$ , and the center-to-center distance as $r = 3.8 \times 10^8\ \text{m}$ . Using $G = 6.67 \times 10^{-11}\ \text{N}\cdot\text{m}^2/\text{kg}^2$ , what is the magnitude of the gravitational force between Earth and the Moon?

$2.0 \times 10^{20}\ \text{N}$ (correct answer)
$7.9 \times 10^{16}\ \text{N}$
$2.0 \times 10^{8}\ \text{N}$
$2.0 \times 10^{28}\ \text{N}$

Explanation: This question tests understanding of Newton's Law of Universal Gravitation and the ability to calculate gravitational force between Earth and the Moon. Newton's Law of Universal Gravitation states that every mass attracts every other mass with a force proportional to the product of their masses and inversely proportional to the square of the distance between their centers: F = G(m₁m₂)/r², where G = 6.67 × 10⁻¹¹ N·m²/kg² is the universal gravitational constant, m₁ and m₂ are the masses in kilograms, r is the center-to-center distance in meters, and F is the attractive force in Newtons. To find the gravitational force between Earth with mass M_E = 6.0 × 10²⁴ kg and the Moon with mass M_Moon = 7.3 × 10²² kg separated by distance r = 3.8 × 10⁸ m, we substitute into F = G(M_E M_Moon)/r²: F = (6.67 × 10⁻¹¹)(6.0 × 10²⁴)(7.3 × 10²²)/(3.8 × 10⁸)² = (6.67 × 10⁻¹¹)(4.38 × 10⁴⁷)/(1.444 × 10¹⁷) = 2.92 × 10³⁷/1.444 × 10¹⁷ = 2.0 × 10²⁰ N. Choice A is correct because it properly applies F = G(M_E M_Moon)/r² with correct astronomical values and accurately handles the scientific notation to get 2.0 × 10²⁰ N. Choice B (7.9 × 10¹⁶ N) has the wrong power of 10, Choice C (2.0 × 10⁸ N) is far too small suggesting an error in powers of 10, and Choice D (2.0 × 10²⁸ N) is far too large, possibly from using r instead of r² in the denominator. When calculating gravitational force between astronomical objects: (1) use the given masses and center-to-center distances carefully, (2) substitute into F = G(m₁m₂)/r² tracking powers of 10 meticulously, (3) expect enormous forces—the 10²⁰ N force between Earth and Moon is what keeps the Moon in orbit and causes Earth's tides. This calculation demonstrates that while gravity seems weak in everyday life, at astronomical scales with massive objects, gravitational forces become the dominant interaction shaping the structure of the solar system and universe.

Question 19

Earth pulls on the Moon with a gravitational force of magnitude $F$ . In this same interaction, what is the magnitude of the gravitational force that the Moon exerts on Earth? (Assume the Earth–Moon distance is $r = 3.8 \times 10^8\ \text{m}$ and use Newton's Law of Universal Gravitation with $G = 6.67 \times 10^{-11}\ \text{N}\cdot\text{m}^2/\text{kg}^2$ .)

$\left(\dfrac{M_{\text{Moon}}}{M_E}\right)F$
$\left(\dfrac{M_E}{M_{\text{Moon}}}\right)F$
$\dfrac{F}{2}$
$F$ (correct answer)

Explanation: This question tests understanding of Newton's Third Law as applied to gravitational forces between Earth and the Moon. Newton's Law of Universal Gravitation states that every mass attracts every other mass with a force proportional to the product of their masses and inversely proportional to the square of the distance between their centers: F = G(m₁m₂)/r², where G = 6.67 × 10⁻¹¹ N·m²/kg² is the universal gravitational constant, m₁ and m₂ are the masses in kilograms, r is the center-to-center distance in meters, and F is the attractive force in Newtons. By Newton's Third Law, forces always come in equal and opposite pairs: if Earth pulls on the Moon with force F, then the Moon must pull on Earth with an equal magnitude force F in the opposite direction, regardless of their different masses. Choice D is correct because Newton's Third Law requires that the forces be equal in magnitude—the force Earth exerts on the Moon equals the force the Moon exerts on Earth, both having magnitude F. Choice A incorrectly suggests the force scales with the mass ratio M_Moon/M_E, making the Moon's force on Earth smaller, Choice B incorrectly suggests the force scales with M_E/M_Moon, making it larger, and Choice C incorrectly suggests the force is halved, all of which violate Newton's Third Law. When analyzing gravitational interactions: (1) forces always come in equal pairs by Newton's Third Law, (2) Earth pulls on Moon with the same magnitude force that Moon pulls on Earth, (3) the accelerations differ because a = F/m, so the Moon accelerates more than Earth due to its smaller mass. Remember that gravitational force is mutual: Earth pulls on you with the same magnitude force that you pull on Earth, regardless of the huge mass difference—the forces are equal but the accelerations are vastly different because a = F/m, so your acceleration toward Earth (9.8 m/s²) is enormous while Earth's acceleration toward you is negligible due to Earth's massive mass.

Question 20

Two identical satellites each have mass $m = 1.0 \times 10^3\ \text{kg}$ . When their centers are $r = 10\ \text{m}$ apart, the gravitational force magnitude is $F$ . If the distance between their centers is doubled to $2r$ , what is the new force magnitude in terms of $F$ ? (Use $F = \dfrac{Gm^2}{r^2}$ .)

$4F$
$2F$
$\dfrac{F}{4}$ (correct answer)
$\dfrac{F}{2}$

Explanation: This question tests understanding of Newton's Law of Universal Gravitation and the inverse square relationship for gravitational force. Newton's Law of Universal Gravitation states that every mass attracts every other mass with a force proportional to the product of their masses and inversely proportional to the square of the distance between their centers: F = G(m₁m₂)/r², where G = 6.67 × 10⁻¹¹ N·m²/kg² is the universal gravitational constant, m₁ and m₂ are the masses in kilograms, r is the center-to-center distance in meters, and F is the attractive force in Newtons. If the distance between two masses changes from r to 2r, the force changes by a factor of (r/2r)² = 1/4, so the new force is F_new = F_original × (r₁/r₂)² = F × (r/2r)² = F × 1/4 = F/4, demonstrating the inverse square relationship. Choice C is correct because it properly applies the inverse square law: when distance doubles, the force becomes 1/4 of the original, so F_new = F/4. Choice A (4F) incorrectly inverts the relationship suggesting force increases when distance increases, Choice B (2F) also wrongly suggests force increases with distance, and Choice D (F/2) incorrectly claims the force only halves when the distance doubles, missing the square in the inverse square law—when distance doubles, force becomes 1/4 (not 1/2). When analyzing how gravitational force changes with distance: (1) remember F ∝ 1/r², so doubling r makes F → F/4, tripling r makes F → F/9, (2) the relationship is inverse square, not just inverse, (3) this rapid decrease with distance makes gravity effectively short-range despite technically extending infinitely. The key insight of the inverse square law is that gravitational force decreases rapidly with distance: double the distance and force drops to 1/4, triple the distance and force drops to 1/9—this r² in the denominator explains why we only feel Earth's gravity strongly and not the gravitational pull of distant mountains or planets.