Suppose the space shuttle is in orbit 390 km from the Earth's surface, and circles the Earth about once every 92.4 minutes. Find the centripetal acceleration of the space shuttle in its orbit. Express your answer in terms of g, the gravitational acceleration at the Earth's surface.
*note my teacher has not taught me anything to do with w = 2pi/T as some help websites have used this route to find the answer, obviously there's another way that my teacher has taught us how to solve this kind of problem and i already know the answer is 0.89g i just don't know how to get there. Thanks.
thank you so much! super helpful i didn't realize i was overlooking the radius of earth!
You're welcome. There are other ways to come up with the same answer using the universal G and the period of the satellite orbit, but this method is quicker.
To find the centripetal acceleration of the space shuttle in its orbit, you can use the formula:
a = v^2 / r
where:
a is the centripetal acceleration,
v is the orbital velocity of the space shuttle, and
r is the radius of the orbit.
In this case, you are given the radius of the orbit (390 km), but you need to find the orbital velocity first. Since the space shuttle completes one orbit in 92.4 minutes, you can determine the velocity using the formula:
v = (2πr) / T
where:
v is the orbital velocity,
r is the radius of the orbit, and
T is the time taken to complete one orbit.
Substituting the given values into the formula, you get:
v = (2π * 390 km) / 92.4 minutes
Now, you need to convert km to meters and minutes to seconds so that all units are consistent. There are 1000 meters in a kilometer and 60 seconds in a minute:
v = (2π * 390,000 m) / (92.4 minutes) * (60 s/1 min) * (1 km / 1000 m)
Once you have the value for v in meters per second, you can substitute it back into the formula for centripetal acceleration:
a = (v^2) / r
Substituting the values, you get:
a = [(2π * 390,000 m) / (92.4 minutes) * (60 s/1 min) * (1 km / 1000 m)]^2 / (390,000 m)
Simplifying the expression gives you the centripetal acceleration in terms of g, the gravitational acceleration at the Earth's surface.
Use the fact that the radius of the Earth is R = 6378 km.
At an altitude of 390 km, the shuttle is R = 6378+390 = 6768 km from the center of the earth. Centripetal a equals gravity g', which is proportional to 1/R^2. Therefore at 6768 km,
g' = g*(6368/6758)^2 = 0.8879 g