(B) The minimum energy required to escape the gravitational field of a planet is equal to the magnitude of the gravitational potential energy of the o

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(B) The minimum energy required to escape the gravitational field of a planet is equal to the magnitude of the gravitational potential energy of the object at the surface: $E = \frac{GMm}{R}$.Given that the densities $\rho$ of the Earth and the Moon are equal,we have $M = \rho V = \rho \cdot \frac{4}{3}\pi R^3$.Since $V_e = 64 V_m$,we have $R_e^3 = 64 R_m^3$,which implies $R_e = 4 R_m$.Also,$M_e = \rho V_e = 64 \rho V_m = 64 M_m$.The energy required is $E = \frac{GMm}{R}$.Therefore,$\frac{E_m}{E_e} = \frac{M_m}{M_e} \cdot \frac{R_e}{R_m} = \frac{1}{64} \cdot \frac{4}{1} = \frac{1}{16}$.Thus,$E_m = \frac{E}{16}$.

Class 11 Physics Gravitation Escape Velocity and Escape Energy

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$A$ rocket has to be launched from the Earth in such a way that it never returns. If $E$ is the minimum energy delivered by the rocket launcher,what should be the minimum energy that the launcher should have if the same rocket is to be launched from the surface of the Moon? Assume that the density of the Earth and the Moon are equal and that the Earth's volume is $64$ times the volume of the Moon.

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A
$\frac{E}{32}$
B
$\frac{E}{16}$
C
$\frac{E}{64}$
D
$\frac{E}{4}$

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Gravitation

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$A$ planet having mass $9 M_e$ and radius $4 R_e$,where $M_e$ and $R_e$ are the mass and radius of the Earth respectively,has an escape velocity in $km/s$ given by: (Given escape velocity on Earth $V_e = 11.2 \times 10^3 \, m/s$)

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