Calculating Work Against Gravity for Orbital Launch

[sour_lemon_1k]

Homework Statement

Materials to construct an 12000 kg lunar observatory are boosted from the Earth's surface to an orbit of 4100 km. In orbit, the observatory is assembled and launched to the moon, 384000 km from Earth.
Compare the work that must be done against Earth's gravity on the two legs of the trip. (Hint: work done against gravity is change in potential energy)

Work to get to orbit: ________ joules
Work to get from orbit to moon: ______________ joules

Homework Equations

possibly U=-GMm/r or U=-GMm/Re + mgh?

Me=5.97*10^24 kg
G=6.637*10^-11
Re=6.37*10^6 m

The Attempt at a Solution

for a)
U=-G(5.97*10^24 kg)(12000 kg)/(6.37*10^6 m+4100*10^3 m)
=-4.54*10^11 joules

(both neg and positive of this answer did not work)

mgh gives you almost the same thing, so i don't think that's the answer either (but feel free to prove me wrong). I've talked with my friend, and she tried -GMm/(Re+alt)-(-GMm/(Re)) and it also didn't work.

didn't try b) because my answers for a were wrong...help?

thanks!

 

[tiny-tim]

Welcome to PF!

Hi sour_lemon_1k ! Welcome to PF!

(try using the X² tag just above the Reply box )

Materials to construct an 12000 kg lunar observatory are boosted from the Earth's surface
…
possibly U=-GMm/r or U=-GMm/Re + mgh?

erm … you've launched it from infinity.

(and it's never "+ mgh" … that only works when h << r )

 

[sour_lemon_1k]

hey thanks for replying! yes it's launched from infinity...but what next! ?__?

also when you say that only works when h<<r, doesn't that apply for all cases? because r = radius of planet + altitude (aka h) so r always > h...unless I am misunderstanding what << means lol.

thanks for helping again!

 

[tiny-tim]

yes it's launched from infinity

No, it's launched from the Earth's surface …

you have to subtract that energy, don't you?​

(my r was the Earth's radius, and my h was the height above it)

 

[sour_lemon_1k]

wait...I GOT IT! thank you!