Calculate Gravitational Force Mathematically - Physics

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 $F = G\dfrac{M_E m}{R_E^2}$ ?
Answer: $5.9 \times 10^{2}\ \text{N}$ 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)/ $R_E$ ² = (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.
A spacecraft of mass $m = 2.0 \times 10^3\ \text{kg}$ is in circular orbit at a distance $r = 1.28 \times 10^7\ \text{m}$ from Earth's center (this is $2R_E$ where $R_E = 6.4 \times 10^6\ \text{m}$ ). Using $M_E = 6.0 \times 10^{24}\ \text{kg}$ and $G = 6.67 \times 10^{-11}\ \text{N}\cdot\text{m}^2/\text{kg}^2$ , what is the magnitude of Earth's gravitational force on the spacecraft?
Answer: $4.9 \times 10^3\ \text{N}$ 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 spacecraft at distance r = 1.28 × 10⁷ m from Earth's center (which is 2 $R_E$ ) has gravitational force F = G( $M_E$ m)/r² = (6.67 × 10⁻¹¹)(6.0 × 10²⁴)(2.0 × 10³)/(1.28 × 10⁷)² = (6.67 × 10⁻¹¹)(1.2 × 10²⁸)/(1.6384 × 10¹⁴) = (8.004 × 10¹⁷)/(1.6384 × 10¹⁴) ≈ 4.9 × 10³ N, which provides the centripetal force for orbit. Choice A is correct because it correctly uses center-to-center distance for orbital calculation and properly applies F = G(m₁m₂)/r² with accurate scientific notation. Choice B is a tempting distractor but fails because it uses r in the denominator instead of r², calculating something like F = GMm/r, which misses the inverse square relationship and makes the force too large by a factor of r. When calculating gravitational force: (1) identify both masses in kg and distance between centers in m, (2) for objects in orbit use r = radius + altitude or given distance, (3) substitute carefully into F = G(m₁m₂)/r² keeping track of scientific notation, (4) check that the result makes sense: at greater distances, force is smaller than at surface. Remember that gravitational force is mutual (Newton's Third Law): Earth pulls on the spacecraft with the same magnitude force that the spacecraft pulls on Earth, regardless of the huge mass difference—the forces are equal but the accelerations are vastly different because a = F/m.
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 $F = \dfrac{Gm^2}{r^2}$ .)
Answer: $\dfrac{F}{4}$ 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.
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 $r = R_E + \text{altitude}$ .)
Answer: $2.2 \times 10^{2}\ \text{N}$ 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.
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)?
Answer: $6.9 \times 10^2\ \text{N}$ 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)/ $R_E$ ² = (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.
A person of mass $m = 70\ \text{kg}$ stands on Earth's surface. Use $F = \dfrac{G M_E m}{R_E^2}$ 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 (weight) on the person?
Answer: $6.8 \times 10^{2}\ \text{N}$ This question tests understanding of Newton's Law of Universal Gravitation and the ability to calculate gravitational force (weight) on a person at Earth's surface. 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)/ $R_E$ ² = (6.67 × 10⁻¹¹)(6.0 × 10²⁴)(70)/(6.4 × 10⁶)² = (6.67 × 10⁻¹¹)(4.2 × 10²⁶)/(4.096 × 10¹³) = 2.80 × 10¹⁶/4.096 × 10¹³ = 6.8 × 10² N = 680 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 calculates the gravitational force as 6.8 × 10² N, which equals the person's weight and verifies that surface gravity g = GM_E/ $R_E$ ² ≈ 9.8 m/s². Choice B (6.8 × 10¹ N) has the wrong power of 10, making the force only 68 N which would give g ≈ 1 m/s², Choice C (6.8 × 10³ N) is too large by a factor of 10, and Choice D (1.1 × 10⁻⁸ N) is absurdly small, possibly from forgetting to include Earth's mass. When calculating gravitational force at a planet's surface: (1) use r = planet radius for the distance from center, (2) the calculated force should equal the object's weight W = mg, (3) this calculation derives the surface gravity g = GM/R² ≈ 9.8 m/s² for Earth. This fundamental calculation shows that the everyday force we call weight is actually the gravitational attraction between our mass and Earth's enormous mass, following the same universal law that governs planetary orbits.
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?
Answer: $2.0 \times 10^{20}\ \text{N}$ 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.
Two objects in space attract each other with gravitational force $F = 1.0 \times 10^{-6}\ \text{N}$ when their centers are separated by $r = 3.0\ \text{m}$ . If the separation is reduced to $1.5\ \text{m}$ (same masses), what is the new gravitational force magnitude? (Use the inverse-square relationship from $F = G\dfrac{m_1 m_2}{r^2}$ .)
Answer: $4.0 \times 10^{-6}\ \text{N}$ This question tests understanding of Newton's Law of Universal Gravitation and the inverse square relationship when distance changes. 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 r₁ = 3.0 m to r₂ = 1.5 m (halving the distance), the force changes by a factor of (r₁/r₂)² = (3.0/1.5)² = 2² = 4, so the new force is F_new = F_original × (r₁/r₂)² = 1.0 × 10⁻⁶ × 4 = 4.0 × 10⁻⁶ N, demonstrating the inverse square relationship. Choice B is correct because it accurately applies the inverse square law showing that when distance halves, force becomes 4 times larger. Choice A (2.5 × 10⁻⁷ N) incorrectly suggests the force decreases when distance decreases, which violates the inverse relationship—objects closer together experience stronger gravitational attraction, not weaker. 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, halve the distance and force increases by 4—this r² in the denominator makes small changes in distance produce large changes in force. Remember that gravitational force is always attractive, so reducing separation always increases the force magnitude.
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 $G = 6.67 \times 10^{-11}\ \text{N}\cdot\text{m}^2/\text{kg}^2$ .)
Answer: $F$ 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.
Earth pulls on the Moon with a gravitational force magnitude given by $F = G\dfrac{M_E M_{\text{Moon}}}{r^2}$ , where $G = 6.67 \times 10^{-11}\,\text{N}\cdot\text{m}^2/\text{kg}^2$ . In this situation, what is the magnitude of the force that the Moon exerts on Earth?
Answer: Equal in magnitude to Earth's force on the Moon. 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 the force is mutual and acts on both objects. By Newton's Third Law, if Earth exerts a gravitational force on the Moon, the Moon must exert an equal magnitude force back on Earth, regardless of their mass difference. The forces form an action-reaction pair: Earth pulls on Moon with force F = G( $M_E$ × M_Moon)/r², and Moon pulls on Earth with the exact same magnitude force F = G(M_Moon × $M_E$ )/r². Choice D is correct because Newton's Third Law requires that the forces be equal in magnitude—the Moon exerts exactly the same magnitude force on Earth that Earth exerts on the Moon. Choice B incorrectly suggests the Moon's force on Earth is smaller because the Moon is less massive, when actually by Newton's Third Law both objects experience equal magnitude forces (though in opposite directions)—the force Earth exerts on the Moon equals the force the Moon exerts on Earth. When analyzing gravitational interactions: (1) forces always come in action-reaction pairs of equal magnitude, (2) the more massive object doesn't exert a larger force, (3) what differs is the acceleration each object experiences (a = F/m), not the force magnitude. 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.
A satellite of mass $m = 1.0 \times 10^3\ \text{kg}$ orbits Earth at an altitude of $h = 4.0 \times 10^5\ \text{m}$ above Earth's surface. Use $M_E = 6.0 \times 10^{24}\ \text{kg}$ , $R_E = 6.4 \times 10^6\ \text{m}$ , and $G = 6.67 \times 10^{-11}\ \text{N}\cdot\text{m}^2/\text{kg}^2$ . Taking $r = R_E + h$ , what is the magnitude of the gravitational force on the satellite?
Answer: $8.7 \times 10^3\ \text{N}$ 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 = 4.0 × 10⁵ m above Earth's surface is at distance r = $R_E$ + h = 6.4 × 10⁶ + 4.0 × 10⁵ = 6.8 × 10⁶ 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)/(6.8 × 10⁶)² = (6.67 × 10⁻¹¹)(6.0 × 10²⁷)/(4.624 × 10¹³) = (4.002 × 10¹⁷)/(4.624 × 10¹³) ≈ 8.7 × 10³ 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 and properly applies F = G(m₁m₂)/r² with accurate scientific notation. Choice C is a tempting distractor but fails because it 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 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: between satellite and Earth are thousands of Newtons. 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.
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?
Answer: $2.0 \times 10^{20}\ \text{N}$ 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.
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 $r = R_E + h$ .)
Answer: $\left(\dfrac{6.8 \times 10^{6}}{7.2 \times 10^{6}}\right)^2 \approx 0.89$ 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.
Two identical metal spheres, each of mass $1.0 \times 10^3\ \text{kg}$ , are floating in deep space with their centers $2.0\ \text{m}$ apart. 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?
Answer: $1.7 \times 10^{-5}\ \text{N}$ This question tests understanding of Newton's Law of Universal Gravitation and the ability to calculate gravitational force between everyday 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. For two identical metal spheres each with mass m = 1.0 × 10³ kg separated by distance r = 2.0 m, we substitute into F = G(m₁m₂)/r²: F = (6.67 × 10⁻¹¹ N·m²/kg²)(1.0 × 10³)(1.0 × 10³)/(2.0)² = (6.67 × 10⁻¹¹)(1.0 × 10⁶)/4.0 = 6.67 × 10⁻⁵/4.0 = 1.67 × 10⁻⁵ ≈ 1.7 × 10⁻⁵ N. Choice A is correct because it properly applies F = G(m₁m₂)/r² with correct values and units, demonstrating that gravitational forces between everyday objects are extremely small. Choice B (1.7 × 10⁻² N) has the calculation correct but uses the wrong power of 10 in scientific notation (10⁻² instead of 10⁻⁵), 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) substitute carefully into F = G(m₁m₂)/r² keeping track of scientific notation, (3) check that the result makes sense: gravitational forces between everyday objects are tiny (~10⁻⁸ to 10⁻⁵ N). The key insight is that while gravity is universal, the small value of G (6.67 × 10⁻¹¹) means gravitational forces between ordinary objects are negligible compared to other forces we experience daily.
Two objects in space have masses $m_1 = 1.0 \times 10^5\,\text{kg}$ and $m_2 = 2.0 \times 10^5\,\text{kg}$ . First they are separated by $r_1 = 1.0 \times 10^3\,\text{m}$ , then by $r_2 = 2.0 \times 10^3\,\text{m}$ . Using $F = G\dfrac{m_1 m_2}{r^2}$ with $G = 6.67 \times 10^{-11}\,\text{N}\cdot\text{m}^2/\text{kg}^2$ , what is the ratio $\dfrac{F_2}{F_1}$ ?
Answer: $\dfrac{1}{4}$ This question tests understanding of how gravitational force changes with distance using the inverse square law. 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. When the distance changes from r₁ = 1.0 × 10³ m to r₂ = 2.0 × 10³ m (doubles), the force ratio is F₂/F₁ = (r₁/r₂)² = (1.0 × 10³/2.0 × 10³)² = (1/2)² = 1/4, demonstrating the inverse square relationship. Choice B is correct because it accurately applies the inverse square law: when distance doubles, the force becomes 1/4 of its original value, so F₂/F₁ = 1/4. Choice A incorrectly claims F₂/F₁ = 1/2, missing the inverse square relationship—this would be true if force were inversely proportional to distance (F ∝ 1/r), but gravitational force follows F ∝ 1/r². When calculating force ratios: (1) use F₂/F₁ = (r₁/r₂)² for the inverse square law, (2) if distance doubles, force becomes 1/4, (3) if distance halves, force becomes 4 times larger. The key insight is that small changes in distance produce large changes in gravitational force due to the square in the denominator—this explains why tides are so sensitive to Moon's distance and why satellites at different altitudes experience very different gravitational forces.
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?
Answer: $2.0 \times 10^{20}\ \text{N}$ 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.
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 $r = R_E$ )?
Answer: $6.9 \times 10^{2}\ \text{N}$ 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)/ $R_E$ ² = (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.
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?
Answer: $2.7 \times 10^{-6}\ \text{N}$ 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.
Two objects in space have masses $m_1 = 1.0 \times 10^{5}\ \text{kg}$ and $m_2 = 1.0 \times 10^{5}\ \text{kg}$ . Their center-to-center separation is reduced from $r = 100\ \text{m}$ to $r = 50\ \text{m}$ . By what factor does the gravitational force change? (Use $F \propto 1/r^2$ .)
Answer: It becomes $4$ times as large. 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 = 100 m to r = 50 m, the force changes by a factor of (100/50)² = 4, so the new force is F_new = F_original × (r₁/r₂)² = [original] × 4, demonstrating the inverse square relationship. Choice C is correct because it accurately applies inverse square law showing force ratio of 4 times larger when distance halves. Choice A 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). 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.
Earth and the Sun are separated by about $r = 1.5 \times 10^{11}\ \text{m}$ (center to center). Use $M_E = 6.0 \times 10^{24}\ \text{kg}$ , $M_{\text{Sun}} = 2.0 \times 10^{30}\ \text{kg}$ , and $G = 6.67 \times 10^{-11}\ \text{N}\cdot\text{m}^2/\text{kg}^2$ . What is the magnitude of the gravitational force between Earth and the Sun?
Answer: $3.6 \times 10^{22}\ \text{N}$ 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 Sun with mass m₂ = 2.0 × 10³⁰ kg separated by distance r = 1.5 × 10¹¹ m, we substitute into F = G(m₁m₂)/r²: F = (6.67 × 10⁻¹¹ N·m²/kg²)(6.0 × 10²⁴ kg)(2.0 × 10³⁰ kg)/(1.5 × 10¹¹ m)² = (6.67 × 10⁻¹¹)(1.2 × 10⁵⁵)/(2.25 × 10²²) = (8.004 × 10⁴⁴)/(2.25 × 10²²) = 3.56 × 10²² N. Choice A is correct because it properly applies F = G(m₁m₂)/r² with correct values and units, handling scientific notation accurately to yield 3.6 × 10²² N. Choice B has the calculation correct but uses the wrong power of 10 in scientific notation (11 instead of 22), 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.

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