Question

A satellite is revolving in a circular orbit at a distance of 2620 km from the surface of the
earth. Calculate
the orbital speed and the period of revolution of the satellite. Radius of the earth
Rę = 6380 km, mass of the earth, M.- 6x 1024 kg and G = 6.67 x 10-11 N m² kg?​

Answer 1

Answer:

2.35 hours and orbital speed is 6.67km/sec

Step-by-step explanation:

given:

h=2626km,Re=6380,

Me.- 6x 1024 kg and G = 6.67 x 10-11 N m² kg?

To find:

the orbital speed and the period of revolution of the satellite=?

now we have to find the orbital velocity,

4v0=square root of(GMe/Re+h)

=square root of((6.67×10−11)(6×1024)/ (6380+2620))

=6.67km/sec

Time period of revolution,

T=2πv0(Re+h)

=6.67×1032×3.14×(6380+2620)

=8474sec

=2.35hours