Question

What is the kinetic energy of a 200 kg satellite as it follows a circular orbit of radius 8x106m around the earth?

Answer 1

Answer:

Kinetic energy of satellite will be 4.98 × 10^9 J.

Explanation:

The velocity V of a satellite orbiting around the earth is given by  

                            V = √(GM/r)    ………. (i)

Where,  

G = gravitational constant = 6.67 × 10^-11 Nm^2/kg^2

M = mass of earth = 5.97 × 10^24 kg

r = radius of orbit = 8 × 10^6 m

By putting values in  equation (i),

                              V = √((6.67 × 10^-11)(5.97 × 10^24)/ 8 × 10^6)    

                              V = 7055.13 m/s

We know that,  

                            Kinetic energy = ½ mV^2

Where,

m = mass of satellite  

so,

                           Kinetic energy = ½ (200)(7055.13)^2

                           Kinetic energy = 4.98 × 10^9 J

Answer 2

Given the equation for the Speed of a Satellite

v = SqRt{Gravitational Constant}{Mass of Earth} divided by the radius given in your problem

we have:


(square root whole term on right side)

v = G Me
———
r


so. (6.67x10^-11)(5.97x10^24)
___________________
(8.0x10^6)


v = 7055 m/s (which is reasonable)


so utilize the Kinetic Energy Formula


KE = 1/2mv^2


KE = 1/2(200)(7055)^2


KE = 4.977x10^9 J