Question 372: Gravitation

Question

The escape velocity of a body on the earth surface is $11.2 \mathrm{~km} / \mathrm{s}$. If the same body is projected upward with velocity $22.4 \mathrm{~km} / \mathrm{s}$, the velocity of this body at infinite distance from the centre of the earth will be:

Accepted Answer

$V_{\infty}=\sqrt{V^2-V_e^2}$
Given than
$\mathrm{V}=2 \mathrm{~V}_e$
So,
$\begin{aligned}
& \mathrm{V}_{\infty}=\sqrt{\left(2 \mathrm{~V}_eight)^2-\mathrm{V}_e^2} \
& \mathrm{~V}_{\infty}=\sqrt{3} \mathrm{~V}_e=11.2 \sqrt{3} \mathrm{~km} / \mathrm{s}\end{aligned}$

Answer

$11.2 \sqrt{2} \mathrm{~km} / \mathrm{s}$

Answer

Zero

Answer

$11.2 \mathrm{~km} / \mathrm{s}$

Answer

$11.2 \sqrt{3} \mathrm{~km} / \mathrm{s}$

Gravitation

The escape velocity of a body on the earth surface is 11.2 \mathrm{~km} / \mathrm{s}. If the same body is projected upward with velocity 22.4 \mathrm{~km} / \mathrm{s}, the velocity of this body at infinite distance from the centre of the earth will be:

Choose the correct answer: