# Work due to gravity

#### ninjagowoowoo

A spacecraft of mass 3120 kg is in a circular orbit a distance 1650 km above the surface of Mars. How much work must the spacecraft engines perform to move the spacecraft to a circular orbit that is 4740 km above the surface?

So I have work = change in energy
so
work= G(M_s)(M_m)(1/R_2 - 1/R_1)

or work = (6.673*10^-11)(3120)(6.41*10^23)(1/(3.4*10^6 + 4740000) - 1/(3.4*10^6 + 1650000)

Does this look right? Or am I missing something stupid?

Related Introductory Physics Homework News on Phys.org

#### Nylex

Yep, method looks right to me. Just remember the units!

#### ninjagowoowoo

For some reason it's still not correct. The only units that I converted were the km -> m. Is there anything else that I needed to convert? OH and my answer should be in joules.

Last edited:

#### OlderDan

Homework Helper
ninjagowoowoo said:
For some reason it's still not correct. The only units that I converted were the km -> m. Is there anything else that I needed to convert? OH and my answer should be in joules.
Looks to me like you have calculated the change in potential energy and set that equal to the work. The work should equal the change in total energy. Total energy is not too hard to derive for a circular orbit. In fact I posted it within the last few days on another thread. See if you can do it yourself, and if not search my posts to find it.

"Work due to gravity"

### Physics Forums Values

We Value Quality
• Topics based on mainstream science
• Proper English grammar and spelling
We Value Civility
• Positive and compassionate attitudes
• Patience while debating
We Value Productivity
• Disciplined to remain on-topic
• Recognition of own weaknesses
• Solo and co-op problem solving