Gravitational Force-Need Help Fast

1. Nov 7, 2007

tibessiba

Gravitational Force---Need Help Fast!!!

1. The problem statement, all variables and given/known data

You are the science officer on a visit to a distant solar system. Prior to landing on a planet you measure its diameter to be 1.8*10^7 m and its rotation period to be 22.3 hours. You have previously determined that the planet orbits from its star with a period of 402 earth days. Once on the surface you find that the acceleration due to gravity is 12.2 m/s^2.

What is the mass of the planet? in kg

What is the mass of the star? in kg

Equations:
Mass(of planet)= g*R^2/G

3. The attempt at a solution

Ok, so I calculated the mass of the planet to be 1.48*10^25 kg using the above equation. M=[(12.2m/s^2)*(9*10^6m)^2]/6.67*10^-11 Nm^2/kg^2.

Now I am stuck with how to find the mass of the star and which equation to use to get it..

Thank you

2. Nov 7, 2007

Anybody???

3. Nov 7, 2007

Astronuc

Staff Emeritus
4. Nov 7, 2007

tibessiba

It states that the planet orbits 2.2*10^11 m from its star. This would them be the distance, correct??

5. Nov 7, 2007

tibessiba

So, then I use T^2=4pi^2/GM somehow??? I guess I'm not exactly sure how...

6. Nov 7, 2007

tibessiba

oops, I forget r^3 in the equation..
so:

T^2=(4pi^2/GM)r^3

7. Nov 7, 2007

tibessiba

T should = 34732800 sec
r should be 2.2*10^11 m

correct??

8. Nov 7, 2007

tibessiba

Using that I calculated the mass of the star to be 1.39 * 10^64...
but that is wrong...
so where did I go wrong??

9. Nov 7, 2007

tibessiba

I calculated it again and this time I got 1.44*10^37, but that is also wrong...

What am I doing wrong?

10. Nov 7, 2007

tibessiba

If I rearrange the equation to find the mass of the star this is what I get:

M=[(2pir^3/2)^2]/GT

Is this right??? It is what I have been using, but I can't come up with the correct answer.

11. Nov 8, 2007

Astronuc

Staff Emeritus
Yes, that is a or r in Kepler's forumula relating period with distance for one mass orbiting another mass, e.g. moon around a planet or planet about a star.

In the simplest form, one may assume that the mass of the star greatly exceeds the mass of the planet.

So $$T^2\,=\,\frac{4{\pi^2}{a^3}}{GM}$$, which can be rearranged to get

$$M\,=\,\frac{4{\pi^2}{a^3}}{GT^2}$$

So substitute in the appropriate numbers

G = 6.67 x 10-11 N m2/kg2, the universal gravitational constant,

a = 2.2 x 1011 m

T = 402 d * 24 h/d * 3600 s/h = 3.47328 x 107 s

Now compare the mass of the star with the mass of the planet. The mass calculated for the star might need adjusting for the mass of the planet since M is the sum of the masses, but if M >> m(planet), the M is approximately the mass of the star.

This is not correct. Be careful about moving terms and exponents. T should be squared and if one brings r^3 inside the parentheses and squares those terms, then one must use r^(3/2) within the parentheses.

Last edited: Nov 8, 2007
12. Nov 8, 2007

tibessiba

Ok... I see.

Thank you very much!!