The escape velocity of a body on the surface of the earth is text11.2kms- 1 . If the earth's mass increases to twice its present value and the radius of the earth becomes half, the escape velocity becomes

Question

The escape velocity of a body on the surface of the earth is text11.2kms- 1 . If the earth's mass increases to twice its present value and the radius of the earth becomes half, the escape velocity becomes (A) 5.6kms- 1 (B) 11.2kms- 1 (C) 22.4kms- 1 (D) 44.8kms- 1

Tardigrade Admin · Accepted Answer

Ve=√(2 G Me/Re)=11.2 kms- 1 Now Mâ²=2Me and Râ²=(Re/2) So Veâ²=√(2 G (2 Me)/((Re)2))=2√(2 G Me/Re)=2× 11.2 = text22.4 kms- 1

Q. The escape velocity of a body on the surface of the earth is $11.2 km s^{- 1}$ . If the earth's mass increases to twice its present value and the radius of the earth becomes half, the escape velocity becomes

$5.6 km s^{- 1}$

$11.2 km s^{- 1}$

$22.4 km s^{- 1}$

$44.8 km s^{- 1}$

$V_{e} = \frac{2 G M _{e}}{R _{e}} = 11.2 km s^{- 1}$
Now $M^{'} = 2 M_{e}$ and $R^{'} = \frac{R _{e}}{2}$
So $V_{e}^{'} = \frac{2 G ( 2 M _{e} )}{( \frac{R _{e}}{2} )} = 2 \frac{2 G M _{e}}{R _{e}} = 2 \times 11.2$
$= 22.4 km s^{- 1}$

Q. The escape velocity of a body on the surface of the earth is 11.2kms−1 . If the earth's mass increases to twice its present value and the radius of the earth becomes half, the escape velocity becomes

5.6kms−1

11.2kms−1

22.4kms−1

44.8kms−1