The Space Shuttle is in a 260.0 mile high orbit above the surface of the earth.

Question

The Space Shuttle is in a 260.0 mile high orbit above the surface of the earth. Draw the free body diagram for the Space Shuttle in orbit. While you might be tempted to draw in a "weight" vector, you should really draw in a "gravitational force of the earth on the shuttle" vector. Assume that the shuttle orbit is a circle. Write out the Newton's Second Law equation for the sum of the forces on the shuttle in the radial direction.

(a) Use your equation to solve for the speed of the Space Shuttle. Take the mass of the earth to be 5.974x1024 kg, and the radius of the earth to be 6.378x106 m.
The speed of the shuttle is: m/s

(b) What is the period of its orbit (in minutes)? HINT: Remember that the period is the time for one full revolution. You know the speed, you can find the distance traveled in one full orbit, so you can find the period.
min

Explanation / Answer

a)

1 Mile =1609.34 m

Radius of the orbit

R=6.378*106+260*1609.34

R=6.8*106 m

acceleration due to gravity at 260 miles

gorbit =g*(r/R)2 =9.8*(6.378*106/6.8*106)2

gorbit=8.63 m/s2

By Conservation of energy

mgorbith =(1/2)mv2

8.63*(260*1609.34) =(1/2)*v2

v=2687.5 m/s

b)

Time period of the orbit

T=2pi*R/V =2pi*(6.8*106)/2687.5

T=265 min

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