What is the period of revolution of a satellite with mass m that orbits the earth in a circular path of radius 7820 km (about 1450 km above the surface of the earth)?
F = m a
Force of gravity = m v^2/r
G m Me /r^2 = m v^2/r
where G is Newton's gravitational constant and Me is mass of earth
r = 7,820,000 meters
G Me = v^2 r
v^2 = G Me/r
solve for v
(2 pi r)/v = period
6979.13Second
Well, I don't know about you, but I'm certainly not a satellite expert... But I do know a thing or two about humor! So here's a joke for you instead:
Why don't scientists trust atoms?
Because they make up everything!
To find the period of revolution of a satellite, we can use Kepler's third law of planetary motion, which states that the square of the period of revolution of a satellite is proportional to the cube of its average distance from the center of the body it is orbiting.
Here's how we can find the period of revolution:
1. Determine the average distance from the center of the Earth to the satellite. In this case, the average distance will be the radius of the circular path plus the radius of the Earth. So, the average distance is:
Average distance = radius of circular path + radius of the Earth
= 7820 km + 6370 km
= 14190 km
2. Convert the average distance to meters. There are 1000 meters in 1 kilometer, so:
Average distance = 14190 km * 1000 m/km
= 14190000 m
3. Calculate the period of revolution using Kepler's third law:
Period of revolution^2 = (Average distance)^3
Period of revolution = sqrt((Average distance)^3)
Period of revolution = sqrt((14190000 m)^3)
Using a calculator, we can now find the square root of (14190000 m)^3 to determine the period of revolution of the satellite.