Work to move a planet's satellite

[lizzyb]

Homework Statement

Calculate the work required to move a planet's satellite of mass 2230 kg from a circular orbit of radius 2R to one of radius 3R, where 5.32 X 10^6 m is the radius of the planet. The mass of the planet is 3.36 X 10^24 kg.

Answer in units of J.

Homework Equations

$$U(r) = -\frac{G M m}{r}$$

The Attempt at a Solution

$$W = \Delta U = U_f - Ui = - \frac{G M_p M_s}{3 R} + \frac{G M_p M_s}{2 R} = - \frac{2 G M_p M_s}{6 R} + \frac{3 G M_p M_s}{6 R} = \frac{G M_p M_s}{6 R}$$

the answer I got (1.5663 X 10^10) was wrong.

 

[a_lawson_2k]

You'd have to integrate dU(r) from 2R to 3R; R varies as the work is done.

 

[lizzyb]

i came up with the same answer via integration

 

[andrevdh]

There will also be a difference in kinetic energy, that is you need to park it in the orbit. Just getting it there is not enough.

 

[lizzyb]

I just used

$$\Delta E = E_f - E_i = - \frac{G M_p M_s}{2}(\frac{1}{R_f} - \frac{1}{R_i})$$

and that was the right answer; so the change in total energy is equal to work?

 

[OlderDan]

I just used

$$\Delta E = E_f - E_i = - \frac{G M_p M_s}{2}(\frac{1}{R_f} - \frac{1}{R_i})$$

and that was the right answer; so the change in total energy is equal to work?

Yes. That is the work-energy principal. The expression you used for the total energy reflects the changes in both potential and kinetic energies between the two orbits. The total energy change is equal to the work done to change the orbit.

 

[andrevdh]

How is this equation derived?

 

[OlderDan]

How is this equation derived?

For a circular orbit

F = mv²/r = GMm/r²

mv² = GMm/r

E = KE + PE = ½mv² - GMm/r = ½GMm/r - GMm/r = -½GMm/r

 

[andrevdh]

Thank you. I thought it would be complicated - integration ...?

 

[OlderDan]

Thank you. I thought it would be complicated - integration ...?

The potential energy function comes from integrating the gravitational force, but that's been done many times so we use the result: PE = -GMm/r