Question:

The total energy of a satellite moving with an orbital velocity $v$ around the earth is

Updated On: Jun 20, 2022
  • $\frac{1}{2}mv^2$
  • $-\frac{1}{2}mv^2$
  • $mv^2$
  • $\frac{3}{2}mv^2$
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The Correct Option is B

Solution and Explanation

Let a satellite is revolving around earth with orbital velocity v. The gravitational potential energy of satellite is
$U=-\frac{GM_em}{R_e}$ ...(i)
where, $M_e$ = mass of earth,
m = mass of satellite

$R_e$ = radius of earth
and G = gravitational constant
The kinetic energy of satellite is
$K=\frac{1}{2}\frac{GM_em}{R_e}$ ...(ii)
Therefore, total energy of satellite, as energy is conserved, is
$E=U+K=\frac{GM_em}{R_e}+\frac{1}{2}\frac{Gm_em}{R_e}$ (iii)
But we know that necessary centripetal force to the satellite is provided by the gravitational force, ie,

$\frac{mv^2}{R_e}=\frac{GM_em}{R_e}$
or $mv^2=\frac{mv^2}=\frac{GM_em}{R_e}$ ......(iv)
Hence, from Eqs. (iii) and (iv), we get
$E=-\frac{1}{2}mv^2$
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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].