Work done by a force changing the distance of a satellite's orbit

[kmb11132]

Homework Statement

A 5530-kg satellite is in a circular Earth orbit that has a radius of 1.81 × 10^7 m. A net external force must act on the satellite to make it change to a circular orbit that has a radius of 8.01 × 10^6 m. What work must the net external force do?

Homework Equations

Orbital Speed:
v = sqrt( G * M_e / r )
where G is the gravitational constant and M_e is Earth's mass

Work-Kinetic Energy Theorem:
W = K_f - K_i = .5*m*v_f^2 - .5*m*v_i^2

The Attempt at a Solution

v_i = sqrt( 6.674e-11 * 5.98e24kg / 1.81e7m )
v_i = 4695.74m/s

v_f = sqrt( 6.674e-11 * 5.98e24kg / 8.01e6m )
v_f = 7058.74m/s

W = .5*m*v_f^2 - .5*m*v_i^2
W = .5*5530kg*7058.74m/s^2 - .5*5530kg*4695.74m/s^2
W = 7.68002e10J

 

[Simon Bridge]

Work = change in energy.

 

[kmb11132]

Work = change in energy.

I incorrectly found the work to be the difference between the initial and final kinetic energies above.

 

[Simon Bridge]

Are there other forms of energy present?

 

[asteriatic]

I would like to point out that the work done by the net external force in changing the satellite's orbit is equivalent to the change in kinetic energy of the satellite. This is because work is defined as the product of force and displacement, and in this case, the force is acting over a distance to change the satellite's orbit. This work is necessary to overcome the gravitational force between the satellite and Earth and change its velocity, resulting in a new circular orbit with a smaller radius. The calculated work of 7.68002e10J is a significant amount and highlights the amount of energy required to change the satellite's orbit. It is also important to note that the work done is directly dependent on the mass of the satellite, as seen in the equation W = .5*m*v_f^2 - .5*m*v_i^2. Therefore, a larger satellite would require more work to change its orbit than a smaller one.