Gravitation Part 1

Given, the time period of the first satellite, T1 = 1h
The time period of the second satellite, T2 = 8 h
The radius of the orbit of nearer satellite, R1 = 2000 kmApplying the Kepler’s third law,
T2∝R3
(

T1

T2

)2=(

R1

R2

)3
⇒

2000

R2

=(

1

8

)

2

3

⇒ R2 = 8000 km
As we know the relation between the angular speed and timeperiod

ω=

θ

T

So, the angular speed of the nearer satellite to the orbit,
ω1=

2π

1

‌rad∕h
and the angular speed of the farther satellite to the orbit,
ω2=

2π

8

=

π

4

‌rad∕h
The speed of the nearer satellite to orbit,
v1=ω1R1=(2π)×2×103‌km∕h
The speed of the farther satellite to the orbit
V2=ω2R2
=(

π

4

)×8000
=π×2×103‌km∕h
Thus, the relative angular speed of the nearer satellite to the farthersatellite,
ω=

V1−V2

R1−R2

=

2π×2×103−π×2×103

8000−2000

=

2π×103

6000

=

π

3

‌rad∕h
Comparing with, ω=

π

x

The value of x = 3.