Question

Assume that a rocket is in Earth’s orbit at 450 km above the surface.

a) Calculate the orbital velocity in km/s

b) Calculate the escape velocity from said orbit and g in this orbit

Answer 1

Let the radius of the earth R = 6380 km = 6.38 × 10⁶ m

Height of the rocket from the earth surface is h = 450 km

= 0.45 × 10⁶ m

Mass of the earth M = 5.98 × 10²⁴ kg

Height of the satellite from the earth's center is r = R + h

= ( 6.38 × 10⁶ + 0.45 × 10⁶ ) m

= 6.83 × 10⁶ m

a) For an orbit at a height ' r ' from the earth's center , orbital velocity is given by

V_o = ( G M / r ) =. { G M / ( R + h ) }

Where G = 6.67 × 10^-11 N m² / kg² is the universal gravitational constant.

V_o = { ( 6.67 × 10^-11 × 5.98 × 10²⁴ ) / ( 6.83 × 10⁶ ) }

=   { ( 39.88 × 10⁷)/ 6.83 }

= ( 58.38 ) × 10³

= 7.640 × 10³ m/s

= 7.64 km/s

So the required orbital velocity is V_o = 7.64 km/s

b) Relation between orbital velocity and escape velocity V_e is

V_e =  2 V_o ( V_e = ( 2 G M / r for the said orbit )

V_e = 1.414 × 7.64 km/s

= 10.80 km/s

So the escape velocity decreases from 11.2 km/s to 10.80 km/s for the said orbit at a height 450 km from the earth's surface.

acceleration due to gravity g' at a height h is given by

g' = g × ( 1 - 2h / R )

= 9.81 × { 1 - ( 2 × 450 / 6380 ) }

= 9.81 × ( 1 - 900 / 6380 )

= 9.81 × ( 1 - 0.1410 )

= 9.81 × 0.859

= 8.426 m/s²

So the acceleration due to gravity at a height h = 450 km is g' = 8.426 m/s²

We can observe that the value of g decreases as we move to higher altitudes.