Question

A satellite, which has a mass of 550kg and a radius of 2.20 meters is orbiting the Earth at an altitude of 375 kilometers.

A) What will be the magnitude of the graviton always force between this satellite and the Earth?

B) What must the velocity of this satellite be in order for the satellite to remain in a stable orbit?

C) What will be the magnitude of the centripetal acceleration of this satellite?

D) How long will it take for this satellite to orbit the earth once

(specific worksheet values attached)

PERIOD N/A 365.25 days 27.3 days 687 days RADIUS 6.96 x 10* m 6.38 x 10 m 1.74 x 10 m 3.39 x 106nm 7.14 x 10' m 1.82 x 10 m 6.00 x 10 m 2.58 x 10 m 2.43 x 10 m 1.90 x 10 m 1.50 x 106 mm A LTİTUDE N/A 1.50 x 10" m 3.80 x 10* m 2.28 x 10"nm 7.79 x 10"m 3.48 x 10* m 1.13 x 101 m 1.22 x 10 m 4.50 x 102 m 3.54 x 10 m 5.91 x 10 m PLANET 1.99 x 103° kg 5.97 x 104 kg 7.35 x 10 kg 6.42 x 1023 kg 1.90 x 10" kg 7.87 x 10 kg 5.68 x 100 kg 1.19 x 10" kg 1.03 x 1026 kg 1.34 x 108 kg 1.50 x 10" kg Jupiter Saturn Titan Neptune Triton Pluto 1.53 x 10 sec 10,759 days 1.42 x 10 sec 60,188 days 5.08 x 103 sec 90,885 days Universal gravitational constant G 6.67 x 101 Nm'/kg? 2 8

Answer 1

Solution:

Mass = m = 550 kg

Radius = 2.20 m

Altitude = 375,000

Radius of the Earth = r= 6.38 x 10⁶ m

Total altitude from earth' center =R= 6.76 x 10⁶ m

A ) Gravitational force = F = G m M / R²

= (6.67 e -11 ) (550) (5.97 e24) / (6.76 e 6)² = 4790 N

B) velocity = $\sqrt{}}$G M / R = $\sqrt{}}$(6.67 e-11)(5.97 e24) / (6.76 e6)

= 7675 m/s

C) centripetal acceleration = v² / R = (7675)² / (6.76 e6)

= 8.71 m s^-2

D) Time period = T = 2 pi R / v = 2 x 3.14 x (6.76 × 10⁶)

= 5534 seconds = 92.2 minutes