The De-orbit Burn

The Shuttle must reduce its velocity at a pre-calculated point in its orbit in order to return to Earth. In order to reduce the velocity and change the orbit of the Shuttle, a maneuver called the de-orbit burn is performed. For this maneuver, the Shuttle is turned in a direction such that the Orbital Maneuvering System (OMS) nozzles point in the direction of the Shuttle's velocity back toward Earth. The OMS engines fire and give the Shuttle a velocity in the opposite direction, thus slowing the spacecraft.

The Shuttle must perform the de-orbit burn to change its orbit so that the perigee, the point in the orbit closest to Earth, is inside of Earth's atmosphere. De-orbit maneuvers are done to lower the perigee of the orbit to 60 miles (or less). An altitude of 60 miles is important because this is where the orbiting spacecraft is recaptured by Earth’s gravity and re-enters Earth’s atmosphere.

Calculate the minimum change in velocity (delta V or ∆V) required for the Space Shuttle to decrease its altitude to 60 miles if it’s orbiting with an apogee of 236 miles and a perigee of 205 miles above the surface of Earth.

Use the rule of thumb that below an altitude of 500 miles, for every 2 feet per second (ft/s) change in the orbiting space craft’s velocity its altitude will change by 1 mile.

and I have to answer in feet per second

I see that someone else is in the VASTS program that didn't understand this problem...

change of altitude: 205-60=145

rule of thumb: deletV=145mile*2/sec.mile

290ft/sec

This is the same question that HAS is using. I'm happy that I'm not the only one who needs help with it.

I am in the WAS program and we have a similar question with different values. How I think it works is much like the first answer. You find the change in perigee needed. Its 205 right now but needs to be 60. So 205-60 is 145 MILES. This is the distance that needs to be changed. Because the rule is 2 ft/s for every mile we do not need to change the units, we have 145 miles and need to change 2 ft/s for each of those miles. So we multiply by 2 and get 290 ft/s. The final thing to consider is that the change is slowing the craft ad not speeding it up so the delta V would be negative and not positive. Making your final answer -290 ft/s.

I hope this helps/is correct and doesn't confuse further.

But I might add that that would be the necessary velocity to do it, but the question asks for the CHANGE in velocity. The change in velocity would be 290ft/sec - (the initial) 2ft/sec, so 288 ft/sec.

isn't that wrong because of the units?

you would have to convert 145 mi to 765,600 ft.

then divide by 2 (because its 2 ft/s) and get 382,800.

then subtract 2 ft/s to get the change, so 382,798.

is that right or did i do something wrong?

I too am in VASTS... that's what I thought. The units would have to be converted first to feet to get an answer in ft/sec.

I am also in the program, but I was wondering did you guys use the program to help you at all?

My computer is really slow and I've only been able to get it to work once...

Some hard stuff homie.