Question

An artificial satellite is in a circular orbit around a planet
of radius r = 2.25 × 10^{3} km at a distance d = 380.0 km
from the planet\'s surface. The period of revolution of the
satellite around the planet is T = 1.15 hours. What is the average
density of the planet?

Answer #1

given r= 2.25 x 10³ km

Using Orbital Period Equation

T²/R³ = (4 x П²)/ (G x M_{central})

|Where,

T = period of the satellite

Mcentral = mass of the central body about which the satellite
orbits

G = Gravitational constant ( which equals 6.673 x 10^{-}¹¹
N m^{2}/kg^{2} )

П = 3.14

Given that

T = 1.15 hours = 4140 seconds

R = ( Radius of planet + distance of satellite from planet's
surface)

= ( 2.25 x 10³) + 380

= 2630 km

= 2630000 meters

M_{central} = (4 x П² x R³)/ (G x T²)

= [4 x (3.14)² x (2630000)³ ] / [ 6.673×10¯¹¹ x (4140)² ]

= 6.273 x 10²³ kg

Volume of planet(V) = 4/3 x П x R³

.'. V = 4/3 x 3.14 x (2630000)³

= 7.616 x 10¹⁹ metre³

Hence, average density of planet = Mass/volume

= M_{central}/ Volume

= (6.273 x 10²³ kg) / (7.616 x 10¹⁹ metre³)

= 8236.607 Kg/m³

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