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.

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.

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.

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) = 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 B forgets to include the gravitational constant G or uses an incorrect value (like 6.67 × 10⁻⁸ instead of 6.67 × 10⁻¹¹), 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.

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.

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.

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.

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.

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.

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.