Question 349: Gravitation

Question

A body is projected from earth's surface with thrice the escape velocity from the surface of the earth. What will be its velocity when it will escape the gravitational pull?

Accepted Answer

Energy required to escape the earth's gravitational field is $$\frac{1}{2} \mathrm{mV}_{\mathrm{e}}^2$$
Energy given to the body is $$=\frac{1}{2} \mathrm{~m}\left(3 \mathrm{~V}_{\mathrm{e}}ight)^2$$
$$=\mathrm{\frac{9}{2} m V_{\mathrm{e}}^2}$$
$$	herefore$$ If $$\mathrm{V}$$ is the velocity of the body when it has escaped from earth's gravitational field then
$$\begin{aligned}
\mathrm{\frac{1}{2}mV^2=\frac{9}{2}mV_e^2-\frac{1}{2}mV_e^2} \ 	herefore \quad \frac{1}{2} \mathrm{mV}^2 & =4 \mathrm{mV}_{\mathrm{e}}^2 \
	herefore \quad \mathrm{V}^2 & =8 \mathrm{~V}_{\mathrm{e}}^2 \
	herefore \quad \mathrm{V} & =2 \sqrt{2} \mathrm{~V}_{\mathrm{e}}
\end{aligned}$$

Answer

$$2 \mathrm{~V}_{\mathrm{e}}$$

Answer

$$4 \mathrm{~V}_{\mathrm{e}}$$

Answer

$$2 \sqrt{2} \mathrm{~V}_{\mathrm{e}}$$

Answer

$$\frac{\mathrm{V}_{\mathrm{e}}}{2}$$

Gravitation

A body is projected from earth's surface with thrice the escape velocity from the surface of the earth. What will be its velocity when it will escape the gravitational pull?

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