# A planet has twice the radius but the mean density is 1/4th as compared to earth. What is the radio of the escape velocity from the earth to that from the planet? Option 1) 3:1 Option 2) 1:2 Option 3) 1:1 Option 4) 2:1

As we learnt in

Energy of satellite -

$K=\frac{1}{2}mV^{2}=\frac{GMm}{2r}$

$U=-\frac{GMm}{r}$

$K\rightarrow Kinetic\: energy$

$U\rightarrow Potential\: energy$

$M\rightarrow mass\: of\: planet$

$m\rightarrow mass\: of\: satellite$

$E=K+U$

$E=-\frac{GMm}{2r}$

$E=Total\: energy$

- wherein

$K=-E$

$U=2E$

$U=-2K$

Escape velocity      $\fn_phv =\sqrt{\frac{2GM}{R}}=\sqrt{\frac{2G. \left(\rho. \frac{4 \pi}{3}R^{3} \right )}{R}}$

$\fn_phv =\sqrt{\frac{8\pi}{3}G\rho R^{2}}$

$\fn_phv \frac{V_{p}}{V_{e}}=\sqrt{\frac{\rho_{p}R_{p}^{2}}{\rho_{e}R_{e}^{2}}}$$\fn_phv =\sqrt{\frac{1}{4}\times 4}=1$

Option 1)

3:1

This is incorrect option

Option 2)

1:2

This is incorrect option

Option 3)

1:1

This is correct option

Option 4)

2:1

This is incorrect option

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