AP Physics 1 MCQ Practice by Topic 2027
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- 1,806 topic MCQs total.
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Gravitational Force Practice
All questions are original. Check each answer for immediate feedback and explanation, then use the lesson link inside the explanation when a concept needs repair.
Two masses remain unchanged while their separation is multiplied by 2. The gravitational force becomes what fraction of its original value?
Correct answer: A. Universal gravitation follows an inverse-square dependence, F_g ∝ 1/r².
Two masses remain unchanged while their separation is multiplied by 3. The gravitational force becomes what fraction of its original value?
Correct answer: A. Universal gravitation follows an inverse-square dependence, F_g ∝ 1/r².
Two masses remain unchanged while their separation is multiplied by 4. The gravitational force becomes what fraction of its original value?
Correct answer: A. Universal gravitation follows an inverse-square dependence, F_g ∝ 1/r².
Two masses remain unchanged while their separation is multiplied by 5. The gravitational force becomes what fraction of its original value?
Correct answer: A. Universal gravitation follows an inverse-square dependence, F_g ∝ 1/r².
Two masses remain unchanged while their separation is multiplied by 6. The gravitational force becomes what fraction of its original value?
Correct answer: A. Universal gravitation follows an inverse-square dependence, F_g ∝ 1/r².
Two masses remain unchanged while their separation is multiplied by 7. The gravitational force becomes what fraction of its original value?
Correct answer: A. Universal gravitation follows an inverse-square dependence, F_g ∝ 1/r².
If one interacting mass is multiplied by 2 while the other mass and separation stay fixed, how does the gravitational force change?
Correct answer: A. F_g = Gm₁m₂/r² is directly proportional to either mass.
If one interacting mass is multiplied by 3 while the other mass and separation stay fixed, how does the gravitational force change?
Correct answer: A. F_g = Gm₁m₂/r² is directly proportional to either mass.
If one interacting mass is multiplied by 4 while the other mass and separation stay fixed, how does the gravitational force change?
Correct answer: A. F_g = Gm₁m₂/r² is directly proportional to either mass.
If one interacting mass is multiplied by 5 while the other mass and separation stay fixed, how does the gravitational force change?
Correct answer: A. F_g = Gm₁m₂/r² is directly proportional to either mass.
If one interacting mass is multiplied by 6 while the other mass and separation stay fixed, how does the gravitational force change?
Correct answer: A. F_g = Gm₁m₂/r² is directly proportional to either mass.
If one interacting mass is multiplied by 7 while the other mass and separation stay fixed, how does the gravitational force change?
Correct answer: A. F_g = Gm₁m₂/r² is directly proportional to either mass.
Near Earth’s surface, what is the approximate weight of a 2 kg object if g = 10 m/s²?
Correct answer: A. Near Earth, gravitational force on a small object is well approximated by F_g = mg.
Near Earth’s surface, what is the approximate weight of a 3 kg object if g = 10 m/s²?
Correct answer: A. Near Earth, gravitational force on a small object is well approximated by F_g = mg.
Near Earth’s surface, what is the approximate weight of a 4 kg object if g = 10 m/s²?
Correct answer: A. Near Earth, gravitational force on a small object is well approximated by F_g = mg.
Near Earth’s surface, what is the approximate weight of a 5 kg object if g = 10 m/s²?
Correct answer: A. Near Earth, gravitational force on a small object is well approximated by F_g = mg.
Near Earth’s surface, what is the approximate weight of a 6 kg object if g = 10 m/s²?
Correct answer: A. Near Earth, gravitational force on a small object is well approximated by F_g = mg.
Near Earth’s surface, what is the approximate weight of a 7 kg object if g = 10 m/s²?
Correct answer: A. Near Earth, gravitational force on a small object is well approximated by F_g = mg.
Two spherical objects have centers separated by 2 m and can be modeled as point masses. Which distance belongs in F_g = Gm₁m₂/r²?
Correct answer: A. The inverse-square law uses the separation between the mass centers in the point/spherical model.
Two spherical objects have centers separated by 3 m and can be modeled as point masses. Which distance belongs in F_g = Gm₁m₂/r²?
Correct answer: A. The inverse-square law uses the separation between the mass centers in the point/spherical model.
Two spherical objects have centers separated by 4 m and can be modeled as point masses. Which distance belongs in F_g = Gm₁m₂/r²?
Correct answer: A. The inverse-square law uses the separation between the mass centers in the point/spherical model.
Two spherical objects have centers separated by 5 m and can be modeled as point masses. Which distance belongs in F_g = Gm₁m₂/r²?
Correct answer: A. The inverse-square law uses the separation between the mass centers in the point/spherical model.
Two spherical objects have centers separated by 6 m and can be modeled as point masses. Which distance belongs in F_g = Gm₁m₂/r²?
Correct answer: A. The inverse-square law uses the separation between the mass centers in the point/spherical model.
Two spherical objects have centers separated by 7 m and can be modeled as point masses. Which distance belongs in F_g = Gm₁m₂/r²?
Correct answer: A. The inverse-square law uses the separation between the mass centers in the point/spherical model.
An astronaut in a circular orbit about 200 km above Earth feels ‘weightless.’ Which statement is correct?
Correct answer: A. Orbital weightlessness is not absence of gravity; it is the shared free-fall condition.
An astronaut in a circular orbit about 250 km above Earth feels ‘weightless.’ Which statement is correct?
Correct answer: A. Orbital weightlessness is not absence of gravity; it is the shared free-fall condition.
An astronaut in a circular orbit about 300 km above Earth feels ‘weightless.’ Which statement is correct?
Correct answer: A. Orbital weightlessness is not absence of gravity; it is the shared free-fall condition.
An astronaut in a circular orbit about 350 km above Earth feels ‘weightless.’ Which statement is correct?
Correct answer: A. Orbital weightlessness is not absence of gravity; it is the shared free-fall condition.
An astronaut in a circular orbit about 400 km above Earth feels ‘weightless.’ Which statement is correct?
Correct answer: A. Orbital weightlessness is not absence of gravity; it is the shared free-fall condition.
An astronaut in a circular orbit about 450 km above Earth feels ‘weightless.’ Which statement is correct?
Correct answer: A. Orbital weightlessness is not absence of gravity; it is the shared free-fall condition.
A 2 kg mass and a 5 kg mass attract each other gravitationally. How do the forces on the two masses compare?
Correct answer: A. The gravitational interaction is also a Newton’s-third-law pair.
A 3 kg mass and a 7 kg mass attract each other gravitationally. How do the forces on the two masses compare?
Correct answer: A. The gravitational interaction is also a Newton’s-third-law pair.
A 4 kg mass and a 9 kg mass attract each other gravitationally. How do the forces on the two masses compare?
Correct answer: A. The gravitational interaction is also a Newton’s-third-law pair.
A 5 kg mass and a 11 kg mass attract each other gravitationally. How do the forces on the two masses compare?
Correct answer: A. The gravitational interaction is also a Newton’s-third-law pair.
A 6 kg mass and a 13 kg mass attract each other gravitationally. How do the forces on the two masses compare?
Correct answer: A. The gravitational interaction is also a Newton’s-third-law pair.
A 7 kg mass and a 15 kg mass attract each other gravitationally. How do the forces on the two masses compare?
Correct answer: A. The gravitational interaction is also a Newton’s-third-law pair.
An object moves through a vertical height change of about 10 m near Earth’s surface. Why is treating g as approximately constant often reasonable?
Correct answer: A. When r changes by a very small fraction of Earth’s radius, GM/r² changes very little.
An object moves through a vertical height change of about 30 m near Earth’s surface. Why is treating g as approximately constant often reasonable?
Correct answer: A. When r changes by a very small fraction of Earth’s radius, GM/r² changes very little.
An object moves through a vertical height change of about 50 m near Earth’s surface. Why is treating g as approximately constant often reasonable?
Correct answer: A. When r changes by a very small fraction of Earth’s radius, GM/r² changes very little.
An object moves through a vertical height change of about 70 m near Earth’s surface. Why is treating g as approximately constant often reasonable?
Correct answer: A. When r changes by a very small fraction of Earth’s radius, GM/r² changes very little.
An object moves through a vertical height change of about 90 m near Earth’s surface. Why is treating g as approximately constant often reasonable?
Correct answer: A. When r changes by a very small fraction of Earth’s radius, GM/r² changes very little.
An object moves through a vertical height change of about 110 m near Earth’s surface. Why is treating g as approximately constant often reasonable?
Correct answer: A. When r changes by a very small fraction of Earth’s radius, GM/r² changes very little.