AP Physics C: Mechanics Practice Quiz: Gravitational Force

Correct Answer: C

Based on Newton's third law and the law of universal gravitation ($F_g = G m_1 m_2 / r^2$), the force that mass m1 exerts on m2 is equal in magnitude and opposite in direction to the force that m2 exerts on m1. The interaction is mutual and the forces are equal in magnitude.

Correct Answer: C

The gravitational force is directly proportional to the product of the two masses ($F_g \propto m_1 m_2$). According to the equation $|\vec{F}_{g}|=G\frac{m_{1}m_{2}}{r^{2}}$, doubling one of the masses will double the resulting gravitational force.

Correct Answer: D

The gravitational force follows an inverse square law with respect to the distance between the centers of the two objects ($F_g \propto 1/r^2$). If the distance r is tripled (3r), the force becomes $1/(3r)^2 = 1/(9r^2)$, which is 1/9 of the original force.

Correct Answer: B

The gravitational field strength |g| is defined as the gravitational force per unit mass, as shown by the equation $|\vec{g}|=\frac{|\vec{F_{g}}|}{m}$. It describes the effect of a source mass (M) on the surrounding space, independent of any test mass (m) placed there.

Correct Answer: B

Newton's shell theorem states that for an object outside a uniform spherical distribution of mass, the net gravitational force exerted on the object is the same as if all the sphere's mass were concentrated at its geometric center. The distance used in the calculation is from center to center (4R).

Correct Answer: C

For an object inside a sphere of uniform density, the net gravitational force is caused only by the mass of the sphere at radii smaller than the object's position. The gravitational effects of the outer spherical shell of mass (from R/2 to R) cancel out to zero.

Correct Answer: D

The original force is $F = G m_1 m_2 / r^2$. The new force is $F' = G (2m_1) m_2 / (2r)^2 = G (2m_1 m_2) / (4r^2) = (2/4) * (G m_1 m_2 / r^2) = (1/2)F$.

Correct Answer: C

The gravitational field strength is given by $|\vec{g}|=G\frac{M}{r^{2}}$. At the surface, the distance from the center is r = R. At an altitude R above the surface, the distance from the center is r = R + R = 2R. Since g is proportional to $1/r^2$, the new field strength will be proportional to $1/(2R)^2 = 1/(4R^2)$, which is 1/4 of the surface field strength.

Correct Answer: C

According to Newton's shell theorem, the net gravitational force exerted by a uniform spherical shell on any object located inside the shell is zero. This is due to the symmetrical pull from all parts of the shell canceling each other out perfectly at the center (and at any other interior point).

Correct Answer: C

The relationship between gravitational force and gravitational field is $|\vec{F_{g}}|=m|\vec{g}|$. Therefore, the force is the product of the astronaut's mass and the local gravitational field strength: Fg = (80 kg)(3.7 m/s²) = 296 N.

Correct Answer: C

Inside the planet (r ≤ R), the gravitational field is due only to the mass within radius r. For a uniform density, this results in g being directly proportional to r (a linear increase). Outside the planet (r > R), the planet acts as a point mass, and the field strength follows an inverse square law, decreasing as 1/r².

Correct Answer: C

We have $g = GM/r^2$. We want to find a new distance, r', for a new mass, M' = 4M, such that the field is the same. So, $g = GM'/(r')^2 = G(4M)/(r')^2$. Setting the two expressions for g equal: $GM/r^2 = G(4M)/(r')^2$. Simplifying gives $1/r^2 = 4/(r')^2$, which leads to $(r')^2 = 4r^2$, and thus r' = 2r.

Correct Answer: D

In Newton's law of universal gravitation, 'r' is the distance separating the centers of mass of the two objects, m1 and m2. This is a crucial detail, especially when dealing with large objects like planets and stars.

Correct Answer: C

Let's write the equations for both forces. F1 = G(M*M)/d² = GM²/d². For Pair 2, F2 = G(2M*2M)/(2d)² = G(4M²)/(4d²) = GM²/d². Therefore, the forces are equal: F2 = F1.

Correct Answer: C

This is a direct application of Newton's third law to gravitation. The force the Earth exerts on the Moon is one part of an action-reaction pair. The other part is the force the Moon exerts on the Earth. These two forces are equal in magnitude and opposite in direction. The equation $F_g = G m_{Earth} m_{Moon} / r^2$ gives the magnitude of the force for both bodies.