Acceleration due to gravity problem.Help!

NewtonJR.215

1. The problem statement, all variables and given/known data
"Determine the acceleration due to gravity on the planet Mercury which has a mass of 3.30*10^23kg and a diameter of 4878 km.

2. Relevant equations
Gmp/rp2
Mp is mass of planet

3. The attempt at a solution
First off I needed to change the diameter into the radius. In my class I have to convert km to m.

4878 km*1000= 4,878,000 m

4,878,000/2= 2,439,000

Next, plugged in.

G=6.67*10^-11

Gmp/rp2
(6.67*10^-11)(3.30*10^23)/2,439,000^2 = 3.70 g on Mercury.

I'm pretty sure this is correct just want to make sure. Thanks for looking it over!

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tiny-tim

Homework Helper
Welcome to PF!

Gmp/rp2
(6.67*10^-11)(3.30*10^23)/2,439,000^2 = 3.70 g on Mercury.

I'm pretty sure this is correct just want to make sure. Thanks for looking it over!
Hi NewtonJR! Welcome to PF! That's a very well-laid out solution … I can see exactly what you're doing!

Yes, that's fine … except I think you don't mean that "g" at the end, do you? btw, as a check (which is often useful to make sure we're not 1000 out!), if we assume that Mercury's density is roughly the same as the Earth's, then mass is proportional to r³, so surface gravity is proportional to r, so should be about 40% of g, = about 4. Yippee! NewtonJR.215

Hi NewtonJR! Welcome to PF! That's a very well-laid out solution … I can see exactly what you're doing!

Yes, that's fine … except I think you don't mean that "g" at the end, do you? btw, as a check (which is often useful to make sure we're not 1000 out!), if we assume that Mercury's density is roughly the same as the Earth's, then mass is proportional to r³, so surface gravity is proportional to r, so should be about 40% of g, = about 4. Yippee! Thanks for the welcome! Thanks for the feedback also. No, I didn't mean to put that "g" at the end. NewtonJR.215

This problem is SOLVED.

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