Results 1 to 3 of 3

Math Help - velocity change

  1. #1
    Newbie
    Joined
    Dec 2008
    Posts
    2

    velocity change

    I've got an issue with a simple sounding physics/calculus problem. It's about a space shuttle.

    De-orbit maneuvers are usually done to lower the perigee of the orbit to 60 miles (or less). The Orbiter is captured and re-enters as it passes into the atmosphere at this altitude.
    There is a change of 1 mile for every 2 feet per second (fps) change in velocity when you are below a 500-mile altitude above the Earth.
    Determine the change in velocity (delta-V) that the Shuttle will need to make if it is at an altitude of 220 miles above the Earth at apogee and 212 miles above the Earth at perigee, and needs to drop the perigee to an altitude of 60 miles.




    I don't even understand what it's asking. I know I need to find a change in velocity. But how do I do that with this information?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    Joined
    Apr 2005
    Posts
    14,977
    Thanks
    1121
    Quote Originally Posted by Fyre View Post
    I've got an issue with a simple sounding physics/calculus problem. It's about a space shuttle.

    De-orbit maneuvers are usually done to lower the perigee of the orbit to 60 miles (or less). The Orbiter is captured and re-enters as it passes into the atmosphere at this altitude.
    There is a change of 1 mile for every 2 feet per second (fps) change in velocity when you are below a 500-mile altitude above the Earth.
    Determine the change in velocity (delta-V) that the Shuttle will need to make if it is at an altitude of 220 miles above the Earth at apogee and 212 miles above the Earth at perigee, and needs to drop the perigee to an altitude of 60 miles.




    I don't even understand what it's asking. I know I need to find a change in velocity. But how do I do that with this information?

    "There is a change of 1 mile for every 2 feet per second (fps) change in velocity when you are below a 500-mile altitude above the Earth."

    To go from 212 to 60 miles is a change of 212- 60= 152 miles. Since there is a 1 mile change for every 2 fps change in velocity, you need to change velocity by 2(152)= 304 fps.

    I used the "212 miles" rather than the "220 miles" because that is given as "perigee" just like the target altitude and it makes more sense to compare the same things. Also, of course, the information is given "to lower the perigee".
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Dec 2008
    Posts
    2

    negative?

    Wouldn't there be a negative in there somewhere too?
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 0
    Last Post: February 23rd 2011, 09:17 AM
  2. Velocity and Rate of Change
    Posted in the Calculus Forum
    Replies: 2
    Last Post: February 15th 2010, 05:48 AM
  3. Replies: 2
    Last Post: September 21st 2009, 02:32 PM
  4. Physics / Rate of Change / Velocity
    Posted in the Calculus Forum
    Replies: 3
    Last Post: May 25th 2009, 08:46 AM
  5. general question about velocity and rates of change
    Posted in the Advanced Applied Math Forum
    Replies: 5
    Last Post: July 22nd 2007, 01:40 PM

Search Tags


/mathhelpforum @mathhelpforum