A 512 kg satellite is in a circular orbit about the Earth at a height above the Earth equal to the Earth's mean radius.
(a) Find the satellite's orbital speed.
(b) Find the period of its revolution.
(c) Find the gravitational force acting on it.
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To find the answers to these questions, we can use the principles of circular motion and the laws of gravitation. Here's how you can calculate each of them:
(a) To find the satellite's orbital speed, we can use the formula for the orbital speed of an object in circular motion:
v = sqrt(GM/r)
Where:
v is the orbital speed
G is the gravitational constant (6.67430 × 10^-11 m³/(kg·s²))
M is the mass of the Earth (5.972 × 10^24 kg)
r is the distance between the satellite and the center of the Earth (equal to the Earth's mean radius, which is approximately 6,371 km or 6,371,000 meters)
Plugging in the values, we have:
v = sqrt((6.67430 × 10^-11 m³/(kg·s²))(5.972 × 10^24 kg)/(6,371,000 m))
v ≈ 7,902 m/s
Therefore, the satellite's orbital speed is approximately 7,902 m/s.
(b) To find the period of the satellite's revolution, we can use the formula for the period of an object in circular orbit:
T = 2πr/v
Where:
T is the period of revolution
r is the distance between the satellite and the center of the Earth (equal to the Earth's mean radius, which is approximately 6,371,000 meters)
v is the orbital speed, which we found to be approximately 7,902 m/s
Plugging in the values, we have:
T = 2π(6,371,000 m)/(7,902 m/s)
T ≈ 5,037 seconds
Therefore, the period of the satellite's revolution is approximately 5,037 seconds.
(c) To find the gravitational force acting on the satellite, we can use the formula for gravitational force:
F = (GMm)/r²
Where:
F is the gravitational force
G is the gravitational constant (6.67430 × 10^-11 m³/(kg·s²))
M is the mass of the Earth (5.972 × 10^24 kg)
m is the mass of the satellite (512 kg)
r is the distance between the satellite and the center of the Earth (equal to the Earth's mean radius, which is approximately 6,371,000 meters)
Plugging in the values, we have:
F = ((6.67430 × 10^-11 m³/(kg·s²))(5.972 × 10^24 kg)(512 kg))/(6,371,000 m)²
F ≈ 2,218,154 N
Therefore, the gravitational force acting on the satellite is approximately 2,218,154 N.