Question:
A body is projected up from the surface of the earth with a velocity equal to $( \frac{3}{4}th )$ of its escape velocity. If $R$ be the radius of earth, the height it reaches is
A body is projected up from the surface of the earth with a velocity equal to $( \frac{3}{4}th )$ of its escape velocity. If $R$ be the radius of earth, the height it reaches is
Updated On: Apr 29, 2024
- $ \frac{3R}{10} $
- $ \frac{9R}{7} $
- $ \frac{8R}{5} $
- $ \frac{9R}{5} $
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The Correct Option is B
Solution and Explanation
If body is projected with velocity $\left(v < v_{e}\right)$,
then height upto which it will rise,
$h=\frac{R}{\frac{v_{e}^{2}}{v^{2}}-1}$
$v=\frac{3}{4} v_{e}$ (given)
$\therefore h=\frac{R}{\frac{v_{e}^{2}}{\frac{9}{16} v_{e}^{2}}-1}$
$=\frac{9}{7} R$
then height upto which it will rise,
$h=\frac{R}{\frac{v_{e}^{2}}{v^{2}}-1}$
$v=\frac{3}{4} v_{e}$ (given)
$\therefore h=\frac{R}{\frac{v_{e}^{2}}{\frac{9}{16} v_{e}^{2}}-1}$
$=\frac{9}{7} R$
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Concepts Used:
Gravitation
In mechanics, the universal force of attraction acting between all matter is known as Gravity, also called gravitation, . It is the weakest known force in nature.
Newton’s Law of Gravitation
According to Newton’s law of gravitation, “Every particle in the universe attracts every other particle with a force whose magnitude is,
- F ∝ (M1M2) . . . . (1)
- (F ∝ 1/r2) . . . . (2)
On combining equations (1) and (2) we get,
F ∝ M1M2/r2
F = G × [M1M2]/r2 . . . . (7)
Or, f(r) = GM1M2/r2
The dimension formula of G is [M-1L3T-2].