1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Tricky Spacecraft Problem

  1. Nov 18, 2007 #1
    1. The problem statement, all variables and given/known data

    The radius of the Earth is about 6.38E3 km. A 7.38E3 N spacecraft travels away from Earth. What is the weight of the spacecraft at the following disatnaces.

    a)6.38E3 km
    b)1.31E4 km

    2. Relevant equations

    F=GMeMship/r^2


    3. The attempt at a solution

    but that won't work because the mass of the ship is in N so I divide it by 9.8 and then the two radiuses need to be added so it will look like this

    F = (6.67E-11)(5.98E24)(753.1)/(6.38E3+6.38E3)^2

    =3E17/1.63E8

    =1840490798

    but that is not the right answer when I punch it into Webassign, and I figure once I get the first one the second one is exactly the same way. So what am I doing wrong?

    Oh and the answers are in Newtons
     
  2. jcsd
  3. Nov 18, 2007 #2

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    For one thing, you'd better put the distances in meters. The same as the units of the other constants in that equation. You can also solve the problem more directly by considering ratios. E.g. for part a) compare the force at R with the force at 2R. You don't need any numbers to do that.
     
    Last edited: Nov 18, 2007
  4. Nov 18, 2007 #3

    D H

    User Avatar
    Staff Emeritus
    Science Advisor

    All you have to do is to compute the force. The units of force are Newtons, and "weight" is defined as the gravitational force.

    You're chief problem here is one of units. Here is a very good practice you should try to get in the habit of using: Always express the units in an equation. For example, don't just use [itex]6.67*10^{-11}[/itex], use the value with its units, [itex]6.67*10^{-11}\mathrm{m}^3/\mathrm{kg}/\mathrm{s}^2[/itex]. The same goes for the masses and distances. You have to make sure the units match.
     
  5. Nov 18, 2007 #4
    so, just in units, (m^3/kg/s^2)(kg)(?)/(km^2). So how do I convert m^3 to km? and what is the missing unit. Other than that I have the equation right?
     
  6. Nov 18, 2007 #5

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    The equation is fine. km=1000*m. So km^3=10^9*m^3. But it's easier just to convert you km radii to m. Easier still to work out the answer using proportions. If the radius changes by a factor of K, by what factor does the force change?
     
  7. Nov 18, 2007 #6
    it would be by K
     
  8. Nov 18, 2007 #7

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    Nooo. What power is r in your equation? If R -> 2R then...
     
  9. Nov 18, 2007 #8
    then it would be multiplied by two and squared
     
  10. Nov 18, 2007 #9

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    If you mean the force is divided by 4, yes. 1/R^2 -> 1/(2R)^2=(1/R^2)/4.
     
  11. Nov 18, 2007 #10
    but which one is R, the radius of the earth added to the radius where the spaceship is at, or what?
     
  12. Nov 18, 2007 #11

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    Think about it. a) says compare the force at radius R (radius of the earth), with the force at radius 2R.
     
  13. Nov 18, 2007 #12
    so I just multiply my answer by the number in a)?
     
  14. Nov 18, 2007 #13

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    I don't know what that means. Gravity is an inverse square law. Doubling distance reduces the force by a factor of four. Etc.
     
  15. Nov 18, 2007 #14
    i didn't know that, but where I am going off track here. Since my equation was right shouldn't I have the right answer? Or is it the units thing that I can't figure out?
     
  16. Nov 18, 2007 #15

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    It's the units thing. Your equation is correct. Just convert the km to m. 6.83E3km=6.83E6m.
     
  17. Nov 18, 2007 #16
    Never mind.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Tricky Spacecraft Problem
  1. Tricky Problem (Replies: 1)

  2. Tricky Torque Problem (Replies: 5)

Loading...