Gravitational Force Inside a Uniform Sphere MCQ Questions and Answers
1. A small test mass is placed at distance r from the centre of a uniform solid sphere of mass M and radius R, where r < R. If the test mass is moved to distance 2r, still inside the sphere, how does the magnitude of the gravitational force on it change?
A. It becomes twice its original value
B. It becomes four times its original value
C. It becomes half its original value
D. It remains unchanged
Correct Answer:
It becomes twice its original value
Explanation:
Inside a uniform sphere, only the mass enclosed within radius r contributes to the net force, and the enclosed mass is proportional to r^3. Therefore, F = GMmr/R^3, so the force is directly proportional to r. Replacing r by 2r doubles the force.
2. A uniform solid sphere has mass 8.0 × 10^24 kg and radius 2.0 × 10^6 m. What is the gravitational force on a 5.0 kg particle located 1.0 × 10^6 m from its centre? Take G = 6.67 × 10^−11 N m^2 kg^−2.
A. 3.34 × 10^2 N, directed toward the centre
B. 6.67 × 10^2 N, directed toward the centre
C. 1.33 × 10^3 N, directed toward the centre
D. 3.34 × 10^2 N, directed away from the centre
Correct Answer:
3.34 × 10^2 N, directed toward the centre
Explanation:
For a point inside a uniform sphere, F = GMmr/R^3. Substitution gives F = (6.67 × 10^−11)(8.0 × 10^24)(5.0)(1.0 × 10^6)/(2.0 × 10^6)^3 = 3.34 × 10^2 N. The force is directed toward the centre.
3. A uniform sphere has density ρ. The magnitude of the gravitational acceleration at an interior point a distance r from its centre is best expressed as
A. 4πGρr
B. 4πGρr/3
C. 4πGρR^3/(3r^2)
D. GρR^3/r^2
Correct Answer:
4πGρr/3
Explanation:
The mass enclosed within radius r is (4π/3)ρr^3. Hence the interior acceleration is g = G[(4π/3)ρr^3]/r^2 = (4πGρ/3)r, directed toward the centre.
4. For a uniform solid sphere of radius R and mass M, at what distance from the centre is the magnitude of the gravitational force on a test mass greatest, considering only points from the centre to the surface?
A. At the centre
B. At R/2
C. At R/√2
D. At the surface, r = R
Correct Answer:
At the surface, r = R
Explanation:
Within the sphere, the force varies as F(r) = GMmr/R^3, which increases linearly with r. Therefore, its maximum value within the material occurs at the surface, where r = R.
5. Two uniform solid spheres have the same total mass M, but their radii are R and 2R. A test mass is placed at the same fractional position r/R inside each sphere, for example at r = R/2 in the first and r = R in the second. What is the ratio of the interior gravitational forces in the first and second spheres?
A. 1:1
B. 2:1
C. 4:1
D. 1:4
Correct Answer:
4:1
Explanation:
At the same fractional position, the force is F = GMm(r/R)/R^2, so for fixed M and fixed r/R it varies as 1/R^2. Doubling the radius reduces the force by a factor of four, giving F_first:F_second = 4:1.
6. A test particle is located inside a uniform sphere at distance r from its centre. Which statement correctly explains why the gravitational force is not determined by the entire mass M using GMm/r^2?
A. The outer layers exert a net outward force that must be subtracted from the inner force
B. The outer spherical shells exert zero net force at the particle, so only the enclosed mass contributes
C. The outer layers repel the particle and cancel part of the inner attraction
D. The gravitational constant becomes smaller inside the sphere
Correct Answer:
The outer spherical shells exert zero net force at the particle, so only the enclosed mass contributes
Explanation:
By the shell theorem, a spherically symmetric shell produces zero net gravitational force at any point inside it. Thus, for a point at radius r, only the mass enclosed within r contributes, giving F = GM(r)m/r^2 rather than GMm/r^2.
7. A uniform sphere of radius R produces a gravitational acceleration g_s at its surface. What is the acceleration at an interior point located at R/3 from the centre?
A. g_s/9
B. g_s/3
C. 3g_s
D. 9g_s
Correct Answer:
g_s/3
Explanation:
For a uniform sphere, the interior gravitational acceleration is proportional to r. Since g_s is the value at r = R, the acceleration at r = R/3 is g_s(R/3)/R = g_s/3.
8. A uniform solid sphere of fixed radius R has its density increased from ρ to 3ρ without changing its size. At any fixed interior distance r from the centre, how does the magnitude of the gravitational force on a given test mass change?
A. It becomes one-third as large
B. It remains unchanged
C. It becomes three times as large
D. It becomes nine times as large
Correct Answer:
It becomes three times as large
Explanation:
Inside a uniform sphere, F = (4πGρm/3)r. At fixed r and fixed test mass, the force is directly proportional to the sphere's density. Increasing ρ to 3ρ therefore triples the force.
9. A particle is released from rest at a point inside a uniform sphere, not at its centre. Neglecting rotation and any other forces, what is the nature of its subsequent radial motion?
A. It moves with constant acceleration toward the centre
B. It executes simple harmonic motion about the centre
C. It moves toward the centre with acceleration increasing as it approaches the centre
D. It remains at rest because the net force inside is zero
Correct Answer:
It executes simple harmonic motion about the centre
Explanation:
Inside a uniform sphere, the gravitational force is F = −(GMm/R^3)r, where the negative sign indicates a restoring force toward the centre. This has the form F = −kr, so the motion is simple harmonic with angular frequency √(GM/R^3).
10. A uniform sphere has surface gravitational acceleration g_s. A test mass is placed at an interior point where the gravitational acceleration is 0.8g_s. What is its distance from the centre?
A. 0.2R
B. 0.4R
C. 0.8R
D. 1.25R
Correct Answer:
0.8R
Explanation:
For an interior point of a uniform sphere, g(r) = g_s(r/R). Setting g(r) = 0.8g_s gives r/R = 0.8, so the point is at r = 0.8R from the centre.
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