Question

How much time a satellite in an orbit at height 35780km above earth's surface would take if the mass of earth would be 4 times its original mass?? Plz help fast..

Answer 1

Here, T is the time period, R is the radius of the orbit, G is the gravitational constant and mass of the earth .
4 × original mass of the earth
= 4 × 5.92 × 10 to the power 24 kg ( Radius of the earth + height of the satellite from the Earth's surface )
= 6400km + 35780km
= 42180 km
=4.2180 × 10 to the power 7
G = 6.67 ×10 to the power -11 -Nm²/kg²
Now,T = 2π√[(4.2180 × 10 to the power 7)³/ (6.67 × 10 to the power -11) × (4 × 5.92 × 10 to the power 24)]
=43309.8 sec
So, T = 43309.8/3600
= 12 hours
Hence, time to complete the revolution by the satellite is 12 hours.

Answer 2

Here, T is the time period, R is the radius of the orbit, G is the gravitational constant and mass of the earth .

radius= 6400km.

4 × original mass of the earth

= 4 × 5.92 × 10 to the power 24 kg ( Radius of the earth + height of the satellite from the Earth's surface )

= 6400km + 35780km

= 42180 km

=4.2180 × 10 to the power 7

G = 6.67 ×10 to the power -11 -Nm²/kg²

Now,T = 2π√[(4.2180 × 10 to the power 7)³/ (6.67 × 10 to the power -11) × (4 × 5.92 × 10 to the power 24)]

=43309.8 sec

So, T = 43309.8/3600

= ~ 12 hours

Hence, time to complete the revolution by the satellite is 12 hours