Solveeit Logo
Login

Question

Question: The escape speed of a body on the earth’s surface is \(11.2 \mathrm {~km} \mathrm {~s} ^ { - 1 }\) ...

The escape speed of a body on the earth’s surface is 11.2 km s−111.2 \mathrm {~km} \mathrm {~s} ^ { - 1 } A body is projected with thrice of this speed. The speed of the body when it escapes the gravitational pull of earth is

A

11.2 km s−111.2 \mathrm {~km} \mathrm {~s} ^ { - 1 }

B

C

22.42 km s−1\frac { 22.4 } { \sqrt { 2 } } \mathrm {~km} \mathrm {~s} ^ { - 1 }

D

22.43 km s−122.4 \sqrt { 3 } \mathrm {~km} \mathrm {~s} ^ { - 1 }

Answer

Explanation

Solution

Let v be the speed of the body when it escapes the gravitational pull of the earth and u be speed of projection of the body from the earth’s surface.

According to law of conservation of mechanical energy.

12mu2−GMEmRE=12mv2−0\frac { 1 } { 2 } \mathrm { mu } ^ { 2 } - \frac { \mathrm { GM } _ { \mathrm { E } } \mathrm { m } } { \mathrm { R } _ { \mathrm { E } } } = \frac { 1 } { 2 } \mathrm { mv } ^ { 2 } - 0

Where m and be masses of the body and earth respectively and is the radius of the earth.

(∵ve=2GMERE)\left( \because \mathrm { v } _ { \mathrm { e } } = \sqrt { \frac { 2 \mathrm { GM } _ { \mathrm { E } } } { \mathrm { R } _ { \mathrm { E } } } } \right)