Question:
At what height above the earth?s surface, the value of $g$ is half of its value on earth?s surface? Given its radius is 6400 km.
At what height above the earth?s surface, the value of $g$ is half of its value on earth?s surface? Given its radius is 6400 km.
Updated On: Jun 7, 2022
- 649.6 km
- 946.4 km
- 2649.6 km
- 624.6 km
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The Correct Option is C
Solution and Explanation
Here g = g/2
But g $=g\left(\frac{R}{R+h}\right)^{^2}$
$\therefore\quad \frac{g}{2}=g\left(\frac{R}{R+h}\right)^{^2}$ or $\left(\frac{R}{R+h}\right)^{^2}=\frac{1}{2}$
or $\quad \frac{R+h}{R}=\sqrt{2}$
or $h=\left(\sqrt{2}-1\right)R = 0.414 R = 0.414 ? 6400$
$= 2649.6$ km.
But g $=g\left(\frac{R}{R+h}\right)^{^2}$
$\therefore\quad \frac{g}{2}=g\left(\frac{R}{R+h}\right)^{^2}$ or $\left(\frac{R}{R+h}\right)^{^2}=\frac{1}{2}$
or $\quad \frac{R+h}{R}=\sqrt{2}$
or $h=\left(\sqrt{2}-1\right)R = 0.414 R = 0.414 ? 6400$
$= 2649.6$ km.
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Concepts Used:
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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.
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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].