Gravitational Force — Practice Problem Solutions

1. $F_g=\dfrac{Gm_1m_2}{r^2}$. $G$ is the universal gravitational constant, $m_1$ and $m_2$ are the two objects' masses, and $r$ is the distance between their centers.

2. The force quadruples (multiplies by $2\times2=4$), since force is proportional to the product of both masses.

3. Since $F_g\propto\dfrac{1}{r^2}$, halving the distance quadruples the force: new force $=4F_g$.

4. $g_{planet}=\dfrac{G(4M_E)}{(2R_E)^2}=\dfrac{4GM_E}{4R_E^2}=\dfrac{GM_E}{R_E^2}=g_{Earth}$. The object's weight on this planet is exactly $W_E$ — unchanged, because the extra mass and the extra radius happen to cancel out perfectly.

5. $F_N=m(g-a)=50(9.8-1.5)=415\text{ N}$.

6. a) $g=\dfrac{GM_{Earth}}{r^2}$. b) At $2r$: $g_{new}=\dfrac{GM}{(2r)^2}=\dfrac{GM}{4r^2}=\dfrac{g}{4}$ — exactly one-fourth its original value, not just "less than."

7. An orbiting astronaut and their spacecraft are both in continuous free fall together, accelerating toward Earth at the same rate under gravity's pull. Because there's no relative force between the astronaut and the spacecraft (no floor pushing up, no seat supporting them), they experience apparent weightlessness — not because gravity has disappeared (it's still nearly full strength at that altitude), but because nothing is providing the normal force that "weight" usually feels like.

8. (C) — more than $mg$, because the elevator's acceleration is directed upward. Moving downward while slowing down means the acceleration points opposite to the (downward) velocity — i.e., upward — and $F_N=mg+ma$ is larger than $mg$ whenever $a$ points up.