A space shuttle is in a circular orbit at a height of 250km, where the acceleration of Earth’s gravity is 93% of its surface value. What is the period of its orbit?
The period of a body orbiting the Earth is defined by T = 2Pisqrt(r^3/µ) where T = the orbit period in minutes, r = the orbital radius in meters and µ = the Earth's gravitational constant of 3.986365x10^14.
I'll let you punch out the numbers.
Given:a = 0.93g
you have to find the radius of the earth which is 6.37*10^6m
R = REarth+ 250km =6.62v 106 m
v=2*pi*r/T a=v^2/R=(2*pi*R/T)^2=4*pi^2R/T
T=sqrt(4*pi^2*R/a
T=sqrt((4*pi^2)(6.62*10^6m) / (.93)(9.8m/s^2)
=5355s
=89min
To find the period of the orbit, we need to use the formula for the period of a circular orbit:
T = 2π√(r³/g)
Where:
T = Period of the orbit
π = Pi, approximately 3.14159
r = Radius of the orbit
g = Acceleration due to gravity
In this case, we are given that the height of the orbit, h, is 250 km above the Earth's surface. So the radius, r, can be calculated as the sum of the radius of the Earth, Re, and the height of the orbit, h:
r = Re + h
The acceleration of Earth's gravity in this orbit, g', can be calculated as:
g' = 93% of the surface value of acceleration due to gravity, g
Now we have all the values we need to calculate the period of the orbit:
Step 1: Calculate the radius of the orbit
r = Re + h
Step 2: Calculate the acceleration due to gravity in this orbit
g' = 0.93 * g
Step 3: Calculate the period of the orbit using the formula
T = 2π√(r³/g')
Let's calculate the period of the orbit using these steps.