At a height of 500km above earth, a satellite will orbit the earth once every 95 minutes. Find the mass of the earth based on this idea.
B) How far above earth's surface would a satellite need to orbit so that it revolves at exactly the same rate as the earth's rotation?
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 = (4π^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) / (4π^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 = (4π^2/G) * r^3 * M
Rearranging the equation, we have:
r = ((T^2 * G) / (4π^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.
To find the mass of the Earth using the given information, we can use Kepler's Third Law and the equation for orbital period.
1) First, let's determine the orbital radius of the satellite. Given that it orbits at a height of 500 km above the Earth's surface, we need to add this distance to the radius of the Earth, which is approximately 6,371 km.
Orbital radius = Height above Earth's surface + Radius of the Earth
= 500 km + 6,371 km
= 6,871 km
2) Now, we can use Kepler's Third Law, which states that the square of the orbital period is proportional to the cube of the orbital radius.
T^2 / r^3 = constant
We have the orbital period T = 95 minutes and the orbital radius r = 6,871 km. Let's convert the period to seconds for consistency.
Orbital period (T) = 95 minutes * 60 seconds per minute
= 5700 seconds
Now we can plug these values into the equation:
(5700 seconds)^2 / (6,871 km)^3 = constant
Solving for the constant, we get:
Constant = (5700 seconds)^2 / (6,871 km)^3
3) Finally, we can use Newton's law of universal gravitation to find the mass of the Earth. The equation is as follows:
F = (G * m1 * m2) / r^2
Where F is the gravitational force between two objects, G is the gravitational constant, m1 and m2 are the masses of the two objects, and r is the distance between them.
In this case, the gravitational force is equal to the centripetal force that keeps the satellite in orbit. Therefore:
F = (m * v^2) / r
Where m is the mass of the satellite and v is the orbital velocity.
Since the satellite travels a circular path, its orbital velocity can be calculated using the formula:
v = (2 * π * r) / T
Plugging in the values, we get:
v = (2 * π * 6,871 km) / 5700 seconds
Now we can equate the centripetal force to the gravitational force and solve for the mass of the Earth (m):
(m * v^2) / r = (G * m * M) / r^2
Where M is the mass of the Earth. Rearranging the equation and canceling out the common factors, we obtain:
M = (v^2 * r) / G
Substituting the values we calculated earlier, we can find the mass of the Earth based on the given information.
Please note that the value of G, the gravitational constant, is approximately 6.674 × 10^(-11) N(m/kg)^2.