The space shuttle is in a 200-mile-high orbit.
What is the shuttle's orbital period?
To determine the space shuttle's orbital period, we can use Kepler's third law of planetary motion, which states that the square of the orbital period is directly proportional to the cube of the average distance between the object and the center of the body it is orbiting.
The formula for Kepler's third law is:
T^2 = (4π^2 / GM) * r^3
Where:
T is the orbital period,
G is the gravitational constant,
M is the mass of the body being orbited,
and r is the average distance between the object and the center of the body.
In this case, we know that the space shuttle is in a 200-mile-high orbit, so the average distance between the shuttle and the center of the Earth is:
r = Earth's radius + altitude of the orbit
= 3959 miles + 200 miles
= 4159 miles
Now, we also need to know the mass of the Earth (M) and the gravitational constant (G) in order to calculate the orbital period (T). The mass of the Earth is approximately 5.972 × 10^24 kilograms, and the gravitational constant is approximately 6.67430 × 10^-11 m³/(kg·s²).
Converting the average distance (r) from miles to meters (1 mile = 1609.34 meters), we get:
r = 4159 miles * 1609.34 meters/mile
= 6,687,005 meters
Now, we can substitute these values into the formula to calculate the orbital period (T):
T^2 = (4π^2 / GM) * r^3
T^2 = (4 * (π^2) / (6.67430 × 10^-11 m³/(kg·s²) * 5.972 × 10^24 kg)) * (6,687,005 meters)^3
Let's now solve this equation to find the value of T.
centripetal acceleration= gravatational acceleration
v^2/r=9.8 (radearth/radearth+200)^2
At this point, you have to convert some units, either to English, or SI
Lets go to SI
v^2=9.8 (radearth)^2/(radearth+3.21E5)
where radius earth is in meters.
V will be in m/sec
Now period is time for one revolution,
v= 2PI (radearth+3.21E5)/T
T= 2PI (radearth+3.21E5)/V
and you can do that.