An Earth satellite moves in a circular orbit 770 km above the Earth's surface. The period of the motion is 100.0 min. What is the speed of the satellite? What is the magnitude of the centripetal acceleration of the satellite?
find the distance from the center of Earth. (re+h) IN METERS
V= distance/time= 2PI*(re+h)/(100*60sec)
solve for v.
centi acceleration: m v^2/(re+h)
To find the speed of the satellite, we can use the formula:
speed = circumference of the orbit / time period
The circumference of the orbit can be found by adding the radius of the orbit to the Earth's radius. The Earth's radius is approximately 6,371 kilometers. So the circumference becomes:
circumference = 2 * π * (radius of the orbit + Earth's radius)
Since the satellite is 770 km above the Earth's surface, the radius of the orbit is the sum of the Earth's radius and the altitude of the satellite:
radius of the orbit = Earth's radius + altitude
Now we can substitute the values into the formula to find the speed of the satellite.
speed = 2 * π * (radius of the orbit + Earth's radius) / time period
speed = 2 * π * (6371 km + 770 km) / (100 min)
Calculating this value will give us the speed of the satellite.
To find the magnitude of the centripetal acceleration of the satellite, we can use the formula:
centripetal acceleration = speed^2 / radius of the orbit
Using the speed and the radius of the orbit, we can calculate the centripetal acceleration.
Now, let's plug in the values and calculate the speed of the satellite and the magnitude of the centripetal acceleration.