Question

a satellite revolves in an orbit close to the surface of a planet of density 6300 kg m^-3. Calculate The time period of the satellite. Take The radius of the planet as 6400 km.

Answer 1

The planet is a sphere of volume 43πR343πR3, and density -

d=M43πR3d=M43πR3

Where MM and RR are the mass and Radius of the planet. Therefore −−

M=d(43πR3)M=d(43πR3) ——— eqn 1

If mm is the mass of the rotating satellite, then the acceleration it feels due to planets gravity -

g=FGmg=FGm, where FGFG is Newton's force of gravity. Thus using Newton's gravitational law -

g=1mGMmR2g=1mGMmR2

Substitute value of MM from eqn 1 here,

g=GR2d(43πR3)g=GR2d(43πR3)

g=4GdπR3g=4GdπR3

This gravitational acceleration acts as the centripetal acceleration for rotating satellite. So -

Centripetal acceleration = gg

ω2R=4GdπR3ω2R=4GdπR3

ω=4Gdπ3−−−−−√ω=4Gdπ3 — — eq 2

Here ωω is the angular velocity of the rotating satellite. Therefore the period of rotation is give as -

T=2πωT=2πω

Substitute ωω from eqn 2 here -

T=2π34Gdπ−−−−−√T=2π34Gdπ

T=3πGd−−−√T=3πGd

T=3∗3.146.67∗10−11∗6300−−−−−−−−−−−−−−−√T=3∗3.146.67∗10−11∗6300

T=9.426.67∗6.3−−−−−−−−√104T=9.426.67∗6.3104

T=0.473469∗104T=0.473469∗104 sec

T=0.473469∗10460∗60T=0.473469∗10460∗60 hours

T=1.315T=1.315 hours for satellite to complete one revolution around the planet.