Question

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

• A. $\frac{1}{2}mv^2$
• B. $-\frac{1}{2}mv^2$
• C. $mv^2$
• D. $\frac{3}{2}mv^2$

Answer 1

Answer: (B)

$-\frac{1}{2}mv^2$

Answer 2

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
$E=-\frac{1}{2}mv^2$