How Does Aerodynamic Drag Influence the Space Shuttle's Orbit Duration?

[quiksilver2871]

Homework Statement

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.

Homework Equations

R=.5DpAv^2

The Attempt at a Solution

Ok i got the portion of (a). i got an answer of 88.4413minutes. that's also with a velocity of 7.7887km/s. I just can't get b. thanks for any help.

 

[Hootenanny]

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

 

[quiksilver2871]

i just don't know how that equation is relevant to the time it takes to drop down to 90% of its speed.

 

[Hootenanny]

What does that equation tell you?

 

[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 don't know how to get the time it takes to drop to the slower speed

 

[Hootenanny]

How about if we write;

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

 

[quiksilver2871]

please forgive me, but i really don't see the connection. I am sry, maybe I am too tired to think straight.

 

[Hootenanny]

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

 

[quiksilver2871]

where did the m come from? If that's the mass of the space shuttle wouldn't it be a really big number to plug in?

 

[Hootenanny]

where did the m come from? If that's the mass of the space shuttle wouldn't it be a really big number to plug in?

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...

 

[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?

 

[Hootenanny]

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?

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.

 

[quiksilver2871]

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

 

[Hootenanny]

Okay, from here;

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

By separation of variables;

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