Gravitational Potential Energy — Practice Problem Solutions

1. $\Delta U_g=mg\Delta y=(0.50)(10)(2.0)=10\text{ J}$. $\Delta U_g=+10\text{ J}$.

2. a) Zero at floor: $U_g=(0.50)(10)(2.0)=10\text{ J}$. b) Zero at table: shelf is $2.0-0.75=1.25\text{ m}$ above the table, so $U_g=(0.50)(10)(1.25)=6.25\text{ J}$. c) Shelf to table top is $\Delta y=-1.25\text{ m}$, so $\Delta U_g=(0.50)(10)(-1.25)=-6.25\text{ J}$ with either zero line (floor: $U$ goes $10\to3.75\text{ J}$; table: $6.25\to0\text{ J}$; both differences are $-6.25\text{ J}$). The zero line changes $U_g$, never $\Delta U_g$.

3. $\Delta y=H-2r$: Sphere 1, $3.0-0.20=2.8\text{ m}$; Sphere 2, $3.0-0.80=2.2\text{ m}$. $|\Delta U_1|=(2.0)(10)(2.8)=56\text{ J}$; $|\Delta U_2|=(2.0)(10)(2.2)=44\text{ J}$. $|\Delta U_1|>|\Delta U_2|$ — the smaller sphere's center of mass falls farther.

4. $U_{t_1}=-\dfrac{GMm}{d}$, $U_{t_2}=-\dfrac{GMm}{2d}$. Ratio $=\dfrac{-1/d}{-1/(2d)}=2$. $U_{t_1}:U_{t_2}=2:1$. $U$ went from $-GMm/d$ to the less negative $-GMm/2d$, so it increased.

5. $U_A=-\dfrac{GMm_0}{r_0}$; $U_B=-\dfrac{GM(4m_0)}{2r_0}=-2\dfrac{GMm_0}{r_0}$. $U_A:U_B=1:2$.

6. $U_i=-\dfrac{GMm}{D}$; $U_f=-\dfrac{GM(m/2)}{2D}=-\dfrac{GMm}{4D}$. $\Delta U=U_f-U_i=-\dfrac{GMm}{4D}+\dfrac{GMm}{D}=\dfrac{3GMm}{4D}$. $\Delta U_g=+\dfrac{3GMm}{4D}$.

7. a) $U(R)=-\dfrac{GMm}{R}$; $U(3R)=-\dfrac{GM(3m)}{3R}=-\dfrac{GMm}{R}$. b) $\Delta U_g=0$. c) Tripling the distance would cut $|U|$ to one-third, but tripling the mass multiplies $|U|$ by 3 — the two factors cancel exactly, so $U$ doesn't change.

8. a) Increases. $U=-Gm_1m_2/r$ is negative; a larger $r$ makes $Gm_1m_2/r$ smaller in magnitude, so $U$ becomes less negative (closer to zero), meaning larger. b) A curve entirely below the $U=0$ axis, rising toward zero as $r$ increases; the larger orbit's point is farther right and higher than the smaller orbit's point.

9. (D) $\Delta U_g=mg\Delta y$ depends only on the drop in height. The object keeps dropping at terminal speed, so $U_g$ keeps decreasing. What stops changing at terminal speed is kinetic energy, not potential energy.

10. Two problems: (1) Gravitational potential energy belongs to the object–Earth system, not to the boulder alone. (2) The number $5000\text{ J}$ depends on the chosen zero line (where $U_g=0$); only changes in $U_g$ have a physical, choice-independent meaning. (A complete answer names both.)

11. a) $U(R)=-\dfrac{GMm}{R}$, $U(3R)=-\dfrac{GMm}{3R}$. $\Delta U=-\dfrac{GMm}{3R}+\dfrac{GMm}{R}=\dfrac{2GMm}{3R}$. $\Delta U_g=+\dfrac{2GMm}{3R}$. b) The student's mistake: because $U=-Gm_1m_2/r$ is negative, a larger distance makes $U$ less negative, so it increases — the satellite gained potential energy moving outward (as it must: work was needed to climb out of the well).