The mass and radius of a satellite is twice that of the earth. If a seconds pendulum is taken to that satellite, what will be its time period?

Explanation

Acceleration due to gravity on the earth.s surface,
`g=(GM)/R^(2)`
Acceleration due to gravity on the surface of the satellite in question,
`g.=(G*2M)/((2R)^(2))=1/2(GM)/R^(2)`
`therefore" "g/(g.)=2`
Time period on the surface of the earth `T=2pisqrt(L/g)`
Time period on the surface of that staellite `T.=2pisqrt(L/(g.))`
Hence, `T/(T.)=sqrt((g.)/g)`
or, `T.=Txxsqrt(g/(g.))=2sqrt2=2.828s" "(becauseT=2s)`.