Answer
a) $T=3.98h$
b) $T=7.31h$
c) The period does not depend on the mass of the satellite.
d) The period is inversely proportional to the square root of the mass of the Earth.
Work Step by Step
(a) We know that
$T=(\frac{2\pi}{\sqrt{GM_E}})(2R_E)^{\frac{3}{2}}$
We plug in the known values to obtain:
$T=(\frac{2\times 3.14}{\sqrt{(6.67\times 10^{-11}N.m^2/Kg^2)(5.97\times 10^{24}Kg)}})(2\times 6.37\times 10^6m)^{\frac{3}{2}}$
This simplifies to:
$T=3.98h$
(b) We know that
$T=(\frac{2\pi}{\sqrt{GM_E}})(3R_E)^{\frac{3}{2}}$
We plug in the known values to obtain:
$T=(\frac{2\times 3.14}{\sqrt{(6.67\times 10^{-11}N.m^2/Kg^2)(5.97\times 10^{24}Kg)}})(3\times 6.37\times 10^6m)^{\frac{3}{2}}$
This simplifies to:
$T=7.31h$
(c) We know that the period does not depend on the mass of the satellite and it only depends on the mass of the Earth and distance from the Earth and the satellite.
(d) We can see from part(a) and part(b) that the period is inversely proportional to the square root of the mass of the Earth.
