Question 429: Gravitation

Question

An object is projected from surface of Earth with a kinetic energy twice that of escape energy ' $K$ ', from surface of Earth. It's kinetic energy when it reaches far away from Earth is

Accepted Answer

If $K=$ escape energy then, $K=\frac{1}{2} m v_e^2=\frac{1}{2} m\left(\sqrt{\frac{2 G M}{R}}ight)^2=\frac{G M m}{R}$
If object is given twice of escape energy or $K E$ then by energy conservation,
$$ \begin{aligned} & U_{	ext {surface }}+K E_{	ext {surface }}=U_{	ext {infinity }}+K E_{	ext {infinity }} \ & \Rightarrow \quad-\frac{G M m}{R}+2 K=0+K E_{	ext {infinity }} \ & \Rightarrow \quad \frac{-G M m}{R}+2 	imes \frac{G M m}{R}=K_{\infty} \ & \Rightarrow \quad K_{\infty}=\frac{G M m}{R}=K_{	ext {escape }} \end{aligned} $$

Answer

K

Answer

$2 K$

Answer

$K / 2$

Answer

0

Gravitation

An object is projected from surface of Earth with a kinetic energy twice that of escape energy ' K ', from surface of Earth. It's kinetic energy when it reaches far away from Earth is

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