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# A body attains a height equal to the radius of the earth when projected from earth surface. The velocity of the body with which it was projected is(A) $\sqrt {\dfrac{{GM}}{R}}$(B) $\sqrt {\dfrac{{2GM}}{R}}$(C) $\sqrt {\dfrac{5}{4}\dfrac{{GM}}{R}}$(D) $\sqrt {\dfrac{{3GM}}{R}}$

Last updated date: 20th Jun 2024
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Hint:In order to solve this problem, first use the concept that at the maximum height point, the velocity of the body becomes zero. After then use the conservation of energy at points A and B. Where point A is the point of earth surface from which body was projected and point B is the maximum height point from which body comes back to earth.

Let the velocity of the body with which it was projected be V and radius of earth is R and mass M.

And given that the height attained by the body is equal to R.
We know that at maximum height point the velocity of the body becomes zero.
Now applying conservation of energy at the surface and at a height $h = R$
$\Rightarrow {(K.E.)_A} + {(P.E.)_A} = {(K.E.)_B} + {(P.E.)_B}$
$\Rightarrow \dfrac{1}{2}m{v^2} + \left( {\dfrac{{ - GMm}}{R}} \right) = 0 + \left( {\dfrac{{ - GMm}}{{2R}}} \right)$
$\Rightarrow \dfrac{1}{2}m{v^2} - \dfrac{{GMm}}{R} = \dfrac{{ - GMm}}{{2R}}$
$\Rightarrow \dfrac{1}{2}m{v^2} = \dfrac{{ - GMm}}{{2R}} + \dfrac{{GMm}}{R}$
$\Rightarrow \dfrac{1}{2}m{v^2} = \dfrac{{GMm}}{{2R}}$
$\Rightarrow {v^2} = \dfrac{{2GMm}}{{2mR}}$
$\Rightarrow {v^2} = \dfrac{{GM}}{R}$
$\therefore v = \sqrt {\dfrac{{GM}}{R}}$
So, the velocity of the body with which it was projected is $\sqrt {\dfrac{{GM}}{R}}$.

Hence, option A is the correct answer.

Note: In order to solve gravitational or satellite problems 2 methods are used.
1. Energy conservation method.
2. Equate centripetal force with the gravitational force.
i.e., $\dfrac{{m{v^2}}}{R} = \dfrac{{GMm}}{{{R^2}}}$