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Acceleration due to gravity problem.Help!

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
Rp is radius 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

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

Science Advisor
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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! :smile:

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? :smile:

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! :biggrin:
 
Hi NewtonJR! Welcome to PF! :smile:

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? :smile:

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! :biggrin:
Thanks for the welcome! Thanks for the feedback also. No, I didn't mean to put that "g" at the end.:cool:
 
This problem is SOLVED.
 

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