Calculating Work for Satellite Orbits Around Earth

[cobrasny]

Homework Statement

A satellite mass 481 is in circular orbit of 9.29x10^6 m above the surface of the Earth
the weight of a 65.1 kg astronaut inside the satellite
What work was done (ignoring friction) to put the satellite into this orbit. Assume it started at rest on the surface of the Earth


Also, A satellite of mass 730 kg is in circular orbit around the Earth, 3.12x10^6 m above the surface of the Earth. Find the minimum work needed to move the satellite outward and put it in circular orbit at a new height of 6.05x10^6 m above the surface of the Earth

Homework Equations

U = -GMm/r
fg = GMem/r^2

The Attempt at a Solution

For the first problem, I tried fg = GMem/r^2, (6.67x10^-11) x (5.97x10^24) x 65.1 / (9.29 x 10^6 + 6.371 x 10^6)^2 and got 105.69 N

For the second, U = -GMm/r; W = Ui-Uf
- (6.67x10^-11) x (5.97x10^24) x 481 / (6.371 x 10^6) - (- (6.67x10^-11) x (5.97x10^24) x 65.1 / (9.29 x 10^6 + 6.371 x 10^6) = -1.78E10 Should the work be negative? Is that the work done by gravity, so the work done to move the satellite is the same magnitude, opposite sign?

I tried a similar strategy for the third problem, and it did not work there either.

Hopefully you can help me understand what I am doing wrong. Thanks in advance.

 

[tiny-tim]

A satellite mass 481 is in circular orbit of 9.29x10^6 m above the surface of the Earth
the weight of a 65.1 kg astronaut inside the satellite
What work was done (ignoring friction) to put the satellite into this orbit. Assume it started at rest on the surface of the Earth
…
U = -GMm/r
fg = GMem/r^2
…
For the first problem, I tried fg = GMem/r^2, (6.67x10^-11) x (5.97x10^24) x 65.1 / (9.29 x 10^6 + 6.371 x 10^6)^2 and got 105.69 N

Hi cobrasny!

Hint: for a circular orbit, the work done is the change in U plus the change in KE.

 

[cobrasny]

So..
- (6.67x10^-11) x (5.97x10^24) x 481 / (9.29 x 10^6 + 6.371 x 10^6) - (- (6.67x10^-11) x (5.97x10^24) x 481 / (6.371 x 10^6) = 1.78E10
is just the change in potential energy.

The change in kinetic energy, 1/2mv^2;
v^2 = GMe/r
Re = 6.371x10^6
v=2pi r/T
Vsurface = 2pi 6.371x10^6/(24hrx60minx60sec)

1/2m (vf^2-vi^2)...
.5 (481) [((6.67x10^-11) x (5.97x10^24) / (9.29 x 10^6 + 6.371 x 10^6)) - (2pi 6.371x10^6/ (24x3600))^2] = 6.063E9

Do I then add the change in Ek and change in Eu? To get 1.78E10 + 6.063E9 = 2.386E10 Joules of work?

 

[cobrasny]

All right, that worked for the second problem.

Two to go:
A satellite of mass 730 kg is in circular orbit around the Earth, 3.12x10^6 m above the surface of the Earth. Find the minimum work needed to move the satellite outward and put it in circular orbit at a new height of 6.05x10^6 m above the surface of the Earth

Delta U = - (6.67x10^-11) x (5.97x10^24) x 730 / (6.05 x 10^6 + 6.371 x 10^6) - (- (6.67x10^-11) x (5.97x10^24) x 730 / (3.12 x 10^6 + 6.371 x 10^6) = 7.2247E9

delta k = .5 (730) [((6.67x10^-11) x (5.97x10^24) / (6.05 x 10^6 + 6.371 x 10^6)) - ((6.67x10^-11) x (5.97x10^24) / (3.12 x 10^6 + 6.371 x 10^6))] = -3.6124E9

So work done = 7.2247E9 + -3.6124E9 = 3.6123E9 Joules (if you could check this I would appreciate it)

A satellite mass 481 is in circular orbit of 9.29x10^6 m above the surface of the Earth
the weight of a 65.1 kg astronaut inside the satellite
I tried fg = GMem/r^2, (6.67x10^-11) x (5.97x10^24) x 65.1 / (9.29 x 10^6 + 6.371 x 10^6)^2 and got 105.69 N
I still don't see what I did wrong here.