Suppose the space shuttle is in orbit 430 km from the Earth's surface, and circles the Earth about once every 93.2 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.
centrip acceleration=w^2 r
where w=2PI/periodinseconds
r is re+altitude
To find the centripetal acceleration of the space shuttle in its orbit, we need to use the formula for centripetal acceleration:
a = (v^2) / r
where a is the centripetal acceleration, v is the orbital velocity, and r is the radius of the orbit.
First, let's find the orbital velocity of the space shuttle. We know that it orbits the Earth once every 93.2 minutes. To convert this to seconds, we multiply by 60:
t = 93.2 minutes * 60 seconds/minute = 5592 seconds
The orbital velocity can then be determined using the formula:
v = 2πr / t
where r is the radius of the orbit and t is the time taken to complete one orbit.
Plugging in the values given in the problem, we have:
v = (2π * 430 km) / 5592 seconds
To simplify the units, convert the radius from kilometers to meters:
r = 430 km * 1000 m/km = 430,000 m
Now we can calculate the orbital velocity:
v = (2π * 430,000 m) / 5592 seconds
Next, we can find the centripetal acceleration using the formula:
a = (v^2) / r
Plugging in the values we found earlier, we have:
a = (v^2) / r
= (v^2) / (430,000 m)
Remember that we need to express the answer in terms of g, the gravitational acceleration at the Earth's surface. The gravitational acceleration at the Earth's surface, g, is approximately 9.8 m/s^2.
Therefore:
a = (v^2) / (430,000 m)
= (2π * 430,000 m / 5592 seconds)^2 / (430,000 m)
≈ (2π)^2 * (430,000 m)^2 / (5592 seconds)^2 / (430,000 m)
≈ (2π)^2 * (430,000 m) / (5592 seconds)^2
≈ (2π)^2 * 9.8 m/s^2
So, the centripetal acceleration of the space shuttle in its orbit is approximately (2π)^2 * 9.8 m/s^2.