Gravitational Fields - is this the right way to solve it?

[fabbo]

I've completed this question and have an answer but I am unsure if my method is correct. The question reads:

A space exploration mission has discovered a new planet with a single moon. The distance between the centres of the planet and its moon is found to be 250000km and the period of rotation of the moon around the planet is 200 hours. On the surface of the planet one experiment shows that an object projected upwards at 20m/s just reaches a height of 14.7m. Taking G to be 6.67 x 10^-11Nm^2/Kg^2 calculate

a) the mass of the planet
I did:

mgh = 1/2 x m x v^2

so g x 14.7 = 1/2 x 20^2

g = 13.6N/kg

I know g = G x (m/r^2) so g is proportional to m

g = Gm so 13.6/6.67 x 10^-11 = m

m = 2.04 x 10^11kg

b) the radius of the planet

i was going to calculate this by T^2 being proportional to r^3.

T = m x r^3

200 x 60 x 60 = 2.04 x 10^11 x r^3

however this gave me an r for the moon as 0.0152m which can't be right...

Is this the right method or have I gone about it in the wrong way?

Any advice would be much appreciated

Thank you

 

[fabbo]

sorry to be a pain but if there's anyone who can help with this question it would be great

 

[Doc Al]

I know g = G x (m/r^2) so g is proportional to m

g = Gm so 13.6/6.67 x 10^-11 = m

How did you get from g = Gm/r^2 to g = Gm ?

 

[Andrew Mason]

I've completed this question and have an answer but I am unsure if my method is correct. The question reads:

A space exploration mission has discovered a new planet with a single moon. The distance between the centres of the planet and its moon is found to be 250000km and the period of rotation of the moon around the planet is 200 hours. On the surface of the planet one experiment shows that an object projected upwards at 20m/s just reaches a height of 14.7m. Taking G to be 6.67 x 10^-11Nm^2/Kg^2 calculate

a) the mass of the planet

b) the radius of the planet

As Doc Al has pointed out, you cannot ignore the r^2 term.

You have to use the period of rotation of the moon around the planet to determine the mass of the planet (and assume that the mass of the planet is much larger than its moon so the orbit radius about the centre of mass is approximately the separation between their centres - otherwise it gets rather more difficult to solve).

Then use the mass of the planet and the acceleration at its surface to determine its surface radius.

AM