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(C) The formula for escape velocity is $v_e = \sqrt{\frac{2GM}{R}}$.Let $M_e$ and $R_e$ be the mass and radius of the Earth,and $M_p$ and $R_p$ be the

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$110$

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(C) The formula for escape velocity is $v_e = \sqrt{\frac{2GM}{R}}$.Let $M_e$ and $R_e$ be the mass and radius of the Earth,and $M_p$ and $R_p$ be the mass and radius of the planet.Given: $M_p = 10 M_e$ and $R_p = \frac{R_e}{10}$.The ratio of escape velocities is given by:$\frac{(v_e)_p}{(v_e)_e} = \sqrt{\frac{M_p}{M_e} 	imes \frac{R_e}{R_p}}$Substituting the given values:$\frac{(v_e)_p}{(v_e)_e} = \sqrt{\frac{10 M_e}{M_e} 	imes \frac{R_e}{R_e/10}} = \sqrt{10 	imes 10} = \sqrt{100} = 10$.Therefore,$(v_e)_p = 10 	imes (v_e)_e = 10 	imes 11 \ km/s = 110 \ km/s$.

$A$ planet in a distant solar system is $10$ times more massive than the earth and its radius is $10$ times smaller. Given that the escape velocity from the earth is $11 \ km/s$,the escape velocity from the surface of the planet would be ........ $km/s$.

Similar Questions

The escape velocity for a body of mass $1\, kg$ from the earth's surface is $11.2\, km/s$. The escape velocity for a body of mass $100\, kg$ would be:

The least velocity required to throw a body away from the surface of a planet so that it may not return is (radius of the planet is $6.4 \times 10^6 \ m$,$g = 9.8 \ m/s^2$).

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$A$ body is projected vertically upwards from the Earth's surface. If the velocity of projection is $\left(\frac{1}{3}\right)$ of the escape velocity,then the height up to which the body rises is $(R = \text{radius of Earth})$

What is the value of escape velocity of a body lying on the surface of Earth and Moon?

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