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Gravitation

Question

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The escape velocity for the earth is 11.2 km/s. The mass of another planet 100 times mass of earth and its radius is 4 times radius of the earth. The escape velocity for the planet is

280 km/s

56.0 km/s

112 km/s

56 km/s

Solution

B.

56.0 km/s

Escape velocity is the minimum speed needed to a free object to escape speed from the gravitational influence of a massive body.

The escape velocity of the planet

V_escape= $\sqrt{\frac{2 G m}{R}} \propto \sqrt{\frac{M}{R}}$

V_escape = $\sqrt{\frac{\frac{2 M_{P}}{R_{P}}}{\frac{2 M_{e}}{R_{e}}}}$

∴ $\frac{V_{P}}{V_{e}} = \sqrt{\frac{M_{P}}{R_{P}} \times \frac{R_{e}}{R_{P}}}$

where V_p - velocity of planet, V_e - velocity of earth

⇒ $\frac{V_{P}}{V_{e}} = \sqrt{100 \times \frac{1}{4}}$

⇒ $\frac{V_{P}}{V_{e}} = 5$

⇒ V_P = 5 × 11.2

⇒ V_P = 56 km/s

Some More Questions From Gravitation Chapter

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Statement – I:
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and
Statement – II
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