# Space Shuttle orbit velocity

1. Oct 8, 2007

### quiksilver2871

1. The problem statement, all variables and given/known data
2. The space shuttle is orbiting the Earth at a distance of about 200 km from its surface.
At that distance, the gravitational acceleration is almost the same as that on the
surface. (a) How long does it take for the shuttle to complete one orbit around the
Earth? Assume that the orbit is circular. (b) The density of air at 200 km is about
5 × 10−10kgr m−3. How many orbits will it take for aerodynamic drag to reduce the
velocity of the shuttle by 10%? The coefficient of aerodynamic drag for the shuttle is
' 0.5, the surface area projected along the direction of motion is ' 400 m2, and the
weight of the shuttle midway through a mission is about one million tons.

2. Relevant equations
R=.5DpAv^2

3. The attempt at a solution
Ok i got the portion of (a). i got an answer of 88.4413minutes. thats also with a velocity of 7.7887km/s. I just cant get b. thanks for any help.

2. Oct 8, 2007

### Hootenanny

Staff Emeritus
Where are you getting stuck with (b)? You seem to have the correct equation (assuming D is the drag coefficient).

3. Oct 8, 2007

### quiksilver2871

i just dont know how that equation is relevant to the time it takes to drop down to 90% of its speed.

4. Oct 8, 2007

### Hootenanny

Staff Emeritus
What does that equation tell you?

5. Oct 8, 2007

### quiksilver2871

it tells me how to solve for resistive force. i know how to solve for the 2 resistive forces at both speeds, but i dont know how to get the time it takes to drop to the slower speed

6. Oct 8, 2007

### Hootenanny

Staff Emeritus

$$F_d = m\frac{dv}{dt}$$

7. Oct 8, 2007

### quiksilver2871

please forgive me, but i really dont see the connection. im sry, maybe im too tired to think straight.

8. Oct 8, 2007

### Hootenanny

Staff Emeritus
Okay, explicitly;

$$-\frac{1}{2}D\rho Av^2 = m\frac{dv}{dt}$$

$$\frac{dv}{dt} + \frac{D\rho A}{2m}v^2 = 0$$

Which is a linear ODE

9. Oct 8, 2007

### quiksilver2871

where did the m come from? If thats the mass of the space shuttle wouldnt it be a really big number to plug in?

10. Oct 8, 2007

### Hootenanny

Staff Emeritus
The m came from Newton's second law (see post #6). And yes m is going to be large, the question says one million tons, but your velocity and the CSA of the shuttle isn't exactly small...

11. Oct 8, 2007

### quiksilver2871

so would i have to convert the weight of it to kilograms or anything? also what would i plug in for v? would i plug in Vf-Vi or what?

12. Oct 8, 2007

### Hootenanny

Staff Emeritus
You should convert all quantities to their SI units and you can't just plug the numbers in, you have to solve the differential equation first.

13. Oct 8, 2007

### quiksilver2871

could you please guide me on what steps i should take to solve this equation. This problem has me completely stumped.

14. Oct 8, 2007

### Hootenanny

Staff Emeritus
Okay, from here;

$$-\frac{1}{2}D\rho Av^2 = m\frac{dv}{dt}$$

By seperation of variables;

$$\int^{0.9v_0}_{v_0} \frac{dv}{v^2} = -\frac{D\rho A}{2m}\int_{0}^{T} dt$$