Acceleration due to gravity problem.Help

[NewtonJR.215]

Homework Statement

"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.

Homework Equations

Gmp/rp2
Mp is mass of planet
Rp is radius of planet

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

2,439,000 is the radius

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!

 

[tiny-tim]

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.