Question

(1) The total energy of a satellite of mass m orbiting with a critical orbital
speed v is
(a) -mv2 (b)-1 2 mv2 (c) 1 2 mv2 (d) mv2​

Answer 1

Answer:

l think option b is the answer

Answer 2

The required "option (b)$-\dfrac{1}{} mv^2$" is correct.

Explanation:

Given,

The satellite of mass = m  and critical orbital  speed = v

We know that,

Total energy of the satellite, $KE=-\dfrac{1}{2} \dfrac{GM_{e}m}{R_{e}}$

Where, $K=\dfrac{1}{2} \dfrac{GM_{e}m}{R_{e}}$

Total energy = Kinetic Energy

$KE=-\dfrac{1}{2}mv^{2}$

Putting the value of KE in the form of mass of a satellite (m) and speed (v).

Hence, the required "option (b)$-\dfrac{1}{} mv^2$" is correct.