Test Index

Gravitation Part 1

Section: Physics

© examsnet.com

Question : 85 of 100

Marks: +1, -0

A planet in a density 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\ \mathrm{km}\ \mathrm{s}^{-1}

, the escape velocity from the surface of the planet would be

$1.1\ \mathrm{km}\ \mathrm{s}^{-1}$
$100\ \mathrm{km}\ \mathrm{s}^{-1}$
$110\ \mathrm{km}\ \mathrm{s}^{-1}$
$0.11\ \mathrm{km}\ \mathrm{s}^{-1}$

Solution:

Let

M_e

is mass of earth then mass of planet

M_p=10 M_e

.

And let

R_e

is radius of earth then radius of planet

R_p = \; \frac{R_c}{10}

Escape velocity of earth,

v_e=\sqrt{\; \frac{2 G M_e}{R_e}}

Escape velocity of planet,

v_p=\sqrt{\; \frac{2 G M_p}{R_p}}

\therefore \; \frac{v_p}{v_e} = \; \frac{\sqrt{\frac{2 G M_p}{R_p}}}{\sqrt{\; \frac{2 G M_e}{R_e}}}

=\sqrt{\; \frac{M_p}{M_e} \times \; \frac{R_e}{R_p}}

=\sqrt{\; \frac{10 M_e}{M_e} \times \; \frac{R_e}{\frac{R_e}{10}}}=10

\therefore v_p = 10 \times v_e

=10 \times 11 = 110\ \mathrm{km/s}

© examsnet.com

Go to Question:

Prev Question

Next Question