To find the mass of the Earth based on a satellite's orbit, we can use Kepler's Third Law of Planetary Motion. This law states that the square of a planet's orbital period (T) is proportional to the cube of the average distance (r) between the planet and the satellite.
Let's first calculate the mass of the Earth using the given information of the satellite orbiting at a height of 500km above the Earth's surface with an orbital period of 95 minutes.
Step 1: Convert the height above the Earth's surface to meters.
Given: Height above Earth's surface = 500 km
1 km = 1000 m
Therefore, Height = 500 km * 1000 m/km = 500,000 m
Step 2: Calculate the average distance (r) between the Earth and the satellite.
Average distance (r) = height above Earth's surface + Earth's radius
Given: Earth's radius = 6,371 km
1 km = 1000 m
Therefore, Earth's radius = 6,371 km * 1000 m/km = 6,371,000 m
Average distance (r) = 500,000 m + 6,371,000 m = 6,871,000 m
Step 3: Convert the orbital period (T) to seconds.
Given: Orbital period (T) = 95 minutes
1 minute = 60 seconds
Therefore, Orbital period (T) = 95 minutes * 60 seconds/minute = 5,700 seconds
Step 4: Use Kepler's Third Law to calculate the mass of the Earth.
Kepler's Third Law: T^2 = (4Ï€^2/G) * r^3 * M
Where T is the orbital period, r is the average distance, G is the gravitational constant (6.67430 x 10^-11 m^3/kg/s^2), and M is the mass of the Earth.
Rearranging the equation, we have:
M = (T^2 * G) / (4Ï€^2 * r^3)
Plugging in the values, we get:
M = (5,700^2 * 6.67430 x 10^-11) / (4 * π^2 * 6,871,000^3)
Calculating this gives us:
M ≈ 5.95 x 10^24 kg
Therefore, the mass of the Earth is approximately 5.95 x 10^24 kg.
Now, let's move on to part B to find out how far above the Earth's surface a satellite would need to orbit to revolve at exactly the same rate as the Earth's rotation.
To match the rotation rate of the Earth, the satellite needs to complete one revolution in 24 hours (since the Earth takes approximately 24 hours to complete one rotation).
Step 1: Convert 24 hours to seconds.
24 hours * 60 minutes/hour * 60 seconds/minute = 86,400 seconds
Step 2: Use Kepler's Third Law to calculate the average distance (r) between the Earth and the satellite.
Kepler's Third Law: T^2 = (4Ï€^2/G) * r^3 * M
Rearranging the equation, we have:
r = ((T^2 * G) / (4Ï€^2 * M))^(1/3)
Plugging in the values, we get:
r = ((86,400^2 * 6.67430 x 10^-11) / (4 * π^2 * 5.95 x 10^24))^(1/3)
Calculating this gives us:
r ≈ 42,163,500 m
Therefore, a satellite would need to orbit approximately 42,163,500 meters above the Earth's surface to revolve at exactly the same rate as the Earth's rotation.