Question:

For a satellite moving in an orbit around the earth, the ratio of kinetic energy to potential energy is

Updated On: Jun 7, 2022
  • 2
  • $ \frac{1}{2} $
  • $ \frac{1}{\sqrt{2}} $
  • $ \sqrt{2} $
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The Correct Option is B

Solution and Explanation

Key Idea Kinetic energy of satellite is half of its potential energy.
Potential energy of satellite
$U=-\frac{G M_{e} m}{R_{e}}$
where $R_{e}$ is radius of earth, $M_{e}$ the mass of earth, $m$ the mass of satellite and $G$ the gravitational constant.
$|U|=\frac{G M_{e} m}{R_{e}}$
Kinetic energy of satellite
$K=\frac{1}{2} \frac{G M_{e} m}{R_{e}}$
Thus, $\frac{K}{|U|}=\frac{1}{2} \frac{G M_{e} m}{R_{e}} \times \frac{R_{e}}{G M_{e} m}=\frac{1}{2}$
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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].