Question:
A body of mass $m$ is taken from the surface of earth to a height equal to twice the radius of earth ($R_e$). The increase in potential energy will be ____ ($g$ is acceleration due to gravity at the surface of earth)
A body of mass $m$ is taken from the surface of earth to a height equal to twice the radius of earth ($R_e$). The increase in potential energy will be ____ ($g$ is acceleration due to gravity at the surface of earth)
Updated On: Apr 12, 2026
- $\frac{1}{2} mgR_e$
- $\frac{3}{4} mgR_e$
- $\frac{1}{4} mgR_e$
- $\frac{2}{3} mgR_e$
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The Correct Option is D
Solution and Explanation
Step 1: Understanding the Concept:
Gravitational potential energy of a mass $m$ at distance $r$ from the center of Earth is $U = -\frac{GM_e m}{r}$. The increase in potential energy is $\Delta U = U_{final} - U_{initial}$.
: Key Formula or Approach:
$\Delta U = -\frac{GM_e m}{R_e + h} - \left(-\frac{GM_e m}{R_e}\right)$.
Given $h = 2R_e$. Also, $g = \frac{GM_e}{R_e^2} \implies GM_e = gR_e^2$.
Step 2: Detailed Explanation:
At the surface ($r = R_e$): $U_i = -\frac{GM_e m}{R_e}$.
At height $h = 2R_e$ ($r = R_e + 2R_e = 3R_e$): $U_f = -\frac{GM_e m}{3R_e}$.
Increase in P.E.:
\[ \Delta U = U_f - U_i = -\frac{GM_e m}{3R_e} + \frac{GM_e m}{R_e} \]
\[ \Delta U = \frac{GM_e m}{R_e} \left(1 - \frac{1}{3}\right) = \frac{GM_e m}{R_e} \left(\frac{2}{3}\right) \]
Substitute $GM_e = gR_e^2$:
\[ \Delta U = \frac{(gR_e^2)m}{R_e} \cdot \frac{2}{3} = \frac{2}{3} mgR_e \]
Step 3: Final Answer:
The increase in potential energy is $\frac{2}{3} mgR_e$.
Gravitational potential energy of a mass $m$ at distance $r$ from the center of Earth is $U = -\frac{GM_e m}{r}$. The increase in potential energy is $\Delta U = U_{final} - U_{initial}$.
: Key Formula or Approach:
$\Delta U = -\frac{GM_e m}{R_e + h} - \left(-\frac{GM_e m}{R_e}\right)$.
Given $h = 2R_e$. Also, $g = \frac{GM_e}{R_e^2} \implies GM_e = gR_e^2$.
Step 2: Detailed Explanation:
At the surface ($r = R_e$): $U_i = -\frac{GM_e m}{R_e}$.
At height $h = 2R_e$ ($r = R_e + 2R_e = 3R_e$): $U_f = -\frac{GM_e m}{3R_e}$.
Increase in P.E.:
\[ \Delta U = U_f - U_i = -\frac{GM_e m}{3R_e} + \frac{GM_e m}{R_e} \]
\[ \Delta U = \frac{GM_e m}{R_e} \left(1 - \frac{1}{3}\right) = \frac{GM_e m}{R_e} \left(\frac{2}{3}\right) \]
Substitute $GM_e = gR_e^2$:
\[ \Delta U = \frac{(gR_e^2)m}{R_e} \cdot \frac{2}{3} = \frac{2}{3} mgR_e \]
Step 3: Final Answer:
The increase in potential energy is $\frac{2}{3} mgR_e$.
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