Question

Draw a diagram and please answer both parts

What was the speed of a space shuttle that orbited Earth at an altitude of 3 x 105 m? Mass of earth = 5.98 x 1024 kg; Radius of earth = 6.38 x 106 m Universal gravitational constant, G = 6.67 x 10-11 N m2 / kg (b) What is its period?

Answer 1

Part A.

Orbit speed of shuttle rotating around earth is given by:

V = sqrt (G*M/R)

M = mass of earth

R = Re + H = 6.38*10^6 + 3*10^5 = 6.68*10^6 m

So,

V = sqrt (6.67*10^-11*5.98*10^24/(6.68*10^6))

V = 7727.3 m/sec = 7.73 km/sec

Part B.

Period of shuttle will be Using Kepler's 3rd law:

T^2 = 4*pi^2*R^3/(G*M)

T = sqrt (4*pi^2*(6.68*10^6)^3/(6.67*10^-11*5.98*10^24))

T = 5431.6 sec = 1.51 hrs

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