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? please solve krke dena step by step​

Answer 1

Answer:

As we the following parameters:-

Mass of earth:-( 6×10^24 kg)

(R1)Radius of earth =6380 km=6380000m

(R2)Distance of satellite from earth:- (2620 km)=2620000m

Gravitation constant (G)=6.67×10^(-11)

Therefore Total Radius:- 6380000+2620000=9000000

So Orbital speed=

$\sqrt{(g \: \times m) \div r }$

=

$\sqrt{((6.67 \times {10}^{ - 11} ) \times 6 \times {10}^{24} ) \div 9000000}$

=

$\sqrt{ {10}^{ 7 \times } \times 4.45 }$

=

$\sqrt{ {10}^{6} } \times \sqrt{44.5}$

=

${10}^{3} \times 6.67$

=6670 m/s^-1

6.67 km/s^-1

HOPE U GOT UR ANSWER