Question 374: Gravitation

Question

A body is thrown from the surface of the earth velocity $$\mathrm{v} / \mathrm{s}$$. The maximum height above the earth's surface upto which it will reach is ($$R=$$ radius of earth, $$g=$$ acceleration due to gravity)

Accepted Answer

At maximum height, the velocity of body becomes zero. Let maximum height be $$h$$.
Applying principle of conservation of energy, we get
$$\begin{aligned}
& \frac{1}{2} m v^2-\frac{G M m}{R}=\frac{1}{2} m(0)^2-\frac{G M m}{(R+h)} \
& 	ext { [where, } m=	ext { mass of body } \
& 	ext { and } M=	ext { mass of earth.] } \
& \Rightarrow \frac{1}{2} m v^2=G M m\left(\frac{1}{R}-\frac{1}{R+h}ight) \
& \Rightarrow \quad \frac{h}{R(R+h)}=\frac{v^2}{2 G M} \
& \Rightarrow \quad \frac{h}{R+h}=\frac{v^2R}{2GM}=\frac{v^2R}{2gR^2} \quad (\because GM=gR^2) \
& \begin{aligned}
& \Rightarrow \quad \frac{R+h}{h} =\frac{2 g R}{v^2} \
& \Rightarrow \quad \frac{R}{h}+1 =\frac{2 g R}{v^2} \
& \Rightarrow \quad \frac{R}{h} =\frac{2 g R}{v^2}-1=\frac{2 g R-v^2}{v^2} \
	ext{or} \quad & h =\frac{v^2 R}{2 g R-v^2}
\end{aligned}
\end{aligned}$$

Answer

$$\frac{v R}{2 g R-v}$$

Answer

$$\frac{2 g A}{v^2(R-1)}$$

Answer

$$\frac{v R^2}{g R-v}$$

Answer

$$\frac{v^2 R}{2 g R-v^2}$$

Gravitation

A body is thrown from the surface of the earth velocity $\mathrm{v} / \mathrm{s}. The maximum height above the earth's surface upto which it will reach is (R= radius of earth, g=$ acceleration due to gravity)

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