# The ratio of escape velocity of earth   to the escape velocity at a planet whose radius and mean density are twice as that of earth is: Option 1) 1 : 2 Option 2) Option 3) 1 : 4 Option 4)

Answers (1)

Escape velocity ( in terms of radius of planet) -

$V_{c}=\sqrt{\frac{2GM}{R}}$

$V_{c}=\sqrt{2gR}$

$V_{c}\rightarrow$ Escape velocity

$R\rightarrow$Radius of earth

- wherein

• depends on the reference body
• greater the value of $\frac{M}{R}$ or $\left ( gR \right )$ greater will be the escape velocity $V_{e}=11.2Km/s$  For earth

escape velocity=$\frac{\sqrt{2GM}}{R}$

$\frac{\sqrt{2G.(\int .\frac{4\pi }{3}R^{3})}}{R}=\sqrt{\frac{8\pi }{3}G\int R^{2}}$

$\frac{V_{p}}{V_{e}}=\frac{\sqrt{\int _{2}.R_{p}^{2}}}{\sqrt{\int }_{e}.R_{e}^{2}}=2\sqrt{2}$

$\frac{V_{e}}{V_{p}}=\frac{1}{2\sqrt{2}}$

Option 1)

1 : 2

This option is incorrect

Option 2)

This option is correct

Option 3)

1 : 4

This option is incorrect

Option 4)

This option is incorrect

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