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7.Gravitation
normal
The escape velocity of a body from earth's surface is $v_e$ . The escape velocity of the same body from a height equal to $R$ from the earth's surface will be
A
$\frac{{{v_e}}}{{\sqrt 2 }}$
B
$\frac{{{v_e}}}{2}$
C
$\frac{{{v_e}}}{{2\sqrt 2 }}$
D
$\frac{{{v_e}}}{4}$
Solution
By energy conservation
$\Rightarrow \frac{1}{2} \mathrm{mv}^{2}=\frac{\mathrm{GMm}}{2 \mathrm{R}} \quad \mathrm{v}_{\mathrm{e}}=\sqrt{\frac{2 \mathrm{GM}}{\mathrm{R}}} \Rightarrow \mathrm{v}=\frac{\mathrm{v}_{\mathrm{e}}}{\sqrt{2}}$
Standard 11
Physics
