Question

A satellite is in a circular orbit around the Earth at an altitude of 1.52 106 m.

(a) Find the period of the orbit. (Hint: Modify Kepler's third law so it is suitable for objects orbiting the Earth rather than the Sun. The radius of the Earth is 6.38 106 m, and the mass of the Earth is 5.98 1024 kg.)

____h

(b) Find the speed of the satellite.

____km/s

(c) Find the acceleration of the satellite.

____m/s2 toward the center of the earth

Answer 1

Solution:

Orbit radius, R = 1.52*10⁶m + 6.38*10⁶m

R = 7.9*10⁶ m

b)The centripetal force is provided by grav. attraction ..

Ms.v²/R = G.Me.Ms / R²

v² = G.Me / R =  (6.67*10^-11) * (5.98*10²⁴) / (7.9*10⁶)

v² = 5.05*10⁷

v = 7.1*10³ m/s = 7.1 km/s

c) Orbit accel. = centripetal accel = v²/R = (7.1*10³)² / (7.9*10⁶)

a = 6.40 m/s²

a) Period, T = 2πR / v

T = 2 *(7.9*10⁶ m) / (7.1*10⁶ m/s)

T = 6.98*10³ s = 1.95 hr

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