Question

A satellite is at an altitude of 852 kilometers above the Earth's surface. The mass of Earth is 5.972 x 10^24 kg and the radius of Earth is 6.378 * 10^6 m.

(1) What is the acceleration due to gravity in the orbit?

(2) What is the orbital velocity and period of the orbit?

(3) Derive the equation for the escape velocity of the orbit and provide the numerical value.

Answer 1

Let the mass of Earth (M) = 5.972 × kg

Radius of Earth (R) = 6.378× m  

Height above the Earth surface (H) = 8.52× m

Gravitational constant (G) = 6.67 ×

Let R+H = D = 7.23 × m

1) Acceleration due to gravity in the orbit a(a)

a = G M /

=6.67× × 5.972× /7.23×

  a = 7.62m

2) A.the orbital velocity ( )   =

   = √(6.67× × 5.972 × /7.23 × )

= 7744.559 m/ s

​​​The orbital velocity is 7744.559 m/s

B.​​​​ Period of the orbit T = (4 /GM)^1/2

on substituting values

T = (4 × × ×/6.67 × × 5.972 × )^1/2

= 6120.184s

  Time period = 6120.184 seconds = 1.7 hours

3) Escape velocity ( ) =

= (2× 9.8 ×7.23 × )^1/2

= 11904.1169 m/s

= 11.904 km/s

  Escape velocity = 11.904 km/s

Derivation

let the minimum velocity to escape from the earth's surface be

Kineticenergy of the object of mass (m) KE = m ​​​​/2

When the projected object is at point P which is at a distance x from the center of the earth.

force of gravity between the object and earth F = G M m /

Work done in taking the body against gravitational attraction from P to Q is given by dW = F dx = GMm / dx

W = dW = GM m / dx   

W = GM m/ D

since K.E = W

1/2×m = GM m/D

we know that g = G M/

on substituting we get

=

Escape velocity =