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

 

Answers (1)
V Vakul

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

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

\frac{V_{p}}{V_{e}}=\sqrt{\frac{\rho_{p}R_{p}^{2}}{\rho_{e}R_{e}^{2}}}=\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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