2.28e7=2*pi*sqrt(7.73e11^3/(6.67e-11*m)

Question

this is my physics homework, dont know how to do it..
full question is: A certain planet is located 7.73 x 10^11 m from its sun. If its orbital period is 2.28 x 10^7 s, find the mass of its sun.

Answer 1

Let the mass of the planet be m and the mass of the sun be M. We can assume that the radius of the planet's orbit is the distance between the centre of the sun and the centre of the planet, that is, the radius of the sun is negligible compared to its distance from the planet.

Gravitational force between the sun and planet = GMm/r²,

where G is the gravitational constant=6.67E-11Nm²/kg² and r=7.73E11m.

The circular motion creates a centripetal force on the planet=mv²/r, where v is the velocity.

These two forces are equal and opposite:

GMm/r²=mv²/r. The length of the orbit is 2πr and the planet takes time T (orbital period) to travel a complete orbit, therefore T=2πr/v, so v=2πr/T. The centripetal force is 4π²rm/T².

GMm/r²=4π²rm/T², M=4π²r³/(GT²).

r³=4.6189E³⁵ or 4.6189×10³⁵ m³. T²=5.1984E14 s².

M=(4π²)(4.6189E35)/(6.67E-11×5.1984E14)=5.259E32kg.

So the mass of the sun is 5.259×10²⁹ metric tonnes.