Question 45147
Newton's law of universal gravitation is goverened by the following equation:
{{{F=G(m[1]m[2])/r^2}}}
where F is the gravitational force fwelt between the two masses m1 and m2,
r is the distance between the centres of the two objects
G is the universal gravitational constant
let m1 be the mass of the planet and m2 be the mass of an object on the surface of that planet.
Now look at Newton's 2nd law:
{{{F=ma}}}
where "a" is an acceleration. If this is gravitational acceleration, we can call it g.
{{{F=mg}}}
also, we can call the mass m1, then 
{{{g=F/m[1]}}} 
If we divide the top equation by m1, we achieve:
{{{g=G(m[2])/r^2}}}

The time period of a simple pendulum is governed by:
{{{T=2*pi*sqrt(L/g)}}}
Now substitute in your expression for g:
{{{T=2*pi*sqrt((L*r^2)/(G*m[2]))}}}
put in your values for r,m2,L and G to find the time period of the pendulum on that planet.
Note G=6.673x10^-11.

I hope this helps.
Adam
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