Question 309: Gravitation

Question

A body is projected vertically upwards from earth's surface of radius '$$R$$' with velocity equal to $$\frac{1^{	ext {rd }}}{3}$$ of escape velocity. The maximum height reached by the body is

Accepted Answer

$$\Delta$$ K.E. $$=\Delta \mathrm{U}$$
Let mass of the particle be $$\mathrm{M}$$ and that of the Earth be $$M_e$$
$$	herefore \quad \frac{1}{2} \mathrm{Mv}^2=\mathrm{GM}_{\mathrm{e}} \mathrm{M}\left(\frac{1}{\mathrm{R}}-\frac{1}{\mathrm{R}+\mathrm{h}}ight)$$ ..... (i)
Also,  $$\mathrm{g}=\frac{\mathrm{GM}_{\mathrm{e}}}{\mathrm{R}^2}$$ ..... (ii)
Equation (i) can be written as,
$$\begin{aligned}
& \frac{1}{2} v^2=G_e\left[\frac{R+h-R}{R(R+h)}ight]=\frac{G_e}{R^2}\left[\frac{R h}{(R+h)}ight] \
& 	herefore \quad \frac{1}{2}\left(\frac{1}{3} v_eight)^2=\frac{g R h}{R+h} \
& 	herefore \quad \frac{1}{2}\left(\frac{1}{3} \sqrt{2 g R}ight)^2=\frac{g R h}{R+h} \
& 	herefore \quad \frac{1}{2} 	imes \frac{1}{9}(2 g R)=\frac{g R h}{R+h} \
& 	herefore \quad \frac{h}{R+h}=\frac{1}{9}
\end{aligned}$$
$$\begin{aligned}
& 	herefore \quad 9 h=R+h \
& 	herefore \quad 8 h=R \
& 	herefore \quad h=\frac{R}{8}
\end{aligned}$$
$$\begin{aligned}
& 	ext { Given } \mathrm{V}=\frac{\mathrm{Ve}}{3} \
& 	herefore \quad \mathrm{h}=\frac{\mathrm{R}}{\left[\frac{\mathrm{V}_e}{\mathrm{~V}_{\mathrm{e}} / 3}ight]^2-1}=\frac{\mathrm{R}}{9-1}=\frac{\mathrm{R}}{8}
\end{aligned}$$

Answer

$$\frac{R}{8}$$

Answer

$$\frac{\mathrm{R}}{6}$$

Answer

$$\frac{\mathrm{R}}{4}$$

Answer

$$\frac{\mathrm{R}}{9}$$

Gravitation

A body is projected vertically upwards from earth's surface of radius '$R' with velocity equal to \frac{1^{\text {rd }}}{3}$ of escape velocity. The maximum height reached by the body is

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