Question:
A satellite of mass $m$ revolves around the earth of radius $R$ at a height $x$ from its surface. If $g$ is the acceleration due to gravity on the surface of the earth, the orbital speed of the satellite is
A satellite of mass $m$ revolves around the earth of radius $R$ at a height $x$ from its surface. If $g$ is the acceleration due to gravity on the surface of the earth, the orbital speed of the satellite is
Updated On: Jul 14, 2022
- $gx$
- $\frac{gR}{R-x}$
- $\frac{gR^{2}}{R+x}$
- $\left(\frac{gR^{2}}{R+x}\right)^{1/ 2}$
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The Correct Option is D
Solution and Explanation
For the satellite, the gravitational force provides the necessary centripetal force i.e.
$\frac{GM_{o}m}{\left(R+X\right)^{2}} = \frac{Mv^{2}_{o}}{\left(R+X\right)}$ and $\frac{GM_{o}}{R^{2}} = g$
$\therefore \, v_{0} =\left(\frac{gR^{2}}{R+x}\right)^{1/ 2}$
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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].