1. Limited time only! Sign up for a free 30min personal 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!

Homework Help: Velocity of earth's orbit?

  1. Apr 10, 2014 #1
    1. The problem statement, all variables and given/known data
    Given that the earth's distance from the sun varies from 1.47 to 1.52x10^11m, determine the minimum and maximum velocities of the Earth in it's orbit around the sun.


    2. Relevant equations

    [itex] F=G\frac{m1m2}{r^2} [/itex]

    [itex] E=K+U [/itex] ???

    3. The attempt at a solution

    I think the way to do this is with K1+U1 = K2+U2 , where one side of the equation is the earth at its closest point to the sun and the other side is the earth at its farthest point. Let Me = mass of earth, Ms = mass of sun, Rn = distance at nearest point, Rf= distance at farthest point, Vn = velocity at nearest point, Vf = velocity at farthest point.

    [itex] K1+U1 = K2 + U2 [/itex]

    [itex] \frac{MeVn^2}{2} + G\frac{MsMe}{Rn} =\frac{MeVf^2}{2} + G\frac{MsMe}{Rf} [/itex]

    the Me's cancel. to solve for Vn replace Vf with [itex] \frac{2piRf}{T} [/itex]

    [itex] \frac{Vn^2}{2} + G\frac{Ms}{Rn} =\frac{2pi^2Rf^2}{T^2} + G\frac{Ms}{Rf} [/itex]

    [itex] \frac{Vn^2}{2} =\frac{2pi^2Rf^2}{T^2} + G\frac{Ms}{Rf} - G\frac{Ms}{Rn} [/itex]


    [itex] Vn^2 = 2(\frac{2pi^2Rf^2}{T^2} + G\frac{Ms}{Rf} - G\frac{Ms}{Rn}) [/itex]

    [itex] Vn = (2(\frac{2pi^2Rf^2}{T^2} + G\frac{Ms}{Rf} - G\frac{Ms}{Rn}))^\frac{1}{2} [/itex]

    then after plugging in I would go back and solve for Vf. Would this give me the correct answer?
     
  2. jcsd
  3. Apr 10, 2014 #2

    BiGyElLoWhAt

    User Avatar
    Gold Member

    well, one thing I want to say is that the earth is revolving, not translating. You don't want 1/2 mv^2, you want 1/2 Iw^2 for your KE, and I'm sure Mr^2 is good enough for I. other than that I think you're on the right track.
     
  4. Apr 10, 2014 #3

    BiGyElLoWhAt

    User Avatar
    Gold Member

    ... I was just looking at your eq. again, and I'm not sure what you're doing with the 2pi*r/T, that is w, but I don't think you're looking at the KE in this situation correctly.
     
  5. Apr 10, 2014 #4
    Ok cool thanks. What does Iw represent?
     
  6. Apr 10, 2014 #5
    I was using the equations for circular motion where v = 2(pi)r/T . I am only vaguely sure of what I am doing though.
     
  7. Apr 10, 2014 #6

    BiGyElLoWhAt

    User Avatar
    Gold Member

    ok, looking again (again)... that'll get you the tangential velocity, but not angular, I'm not sure what you're after exactly.

    Also, I think it's good practice to start with the angular and convert over, seeing as you have KE as a result of rotation, not translation. I think that will get you the correct result, though (this time).
     
  8. Apr 10, 2014 #7

    BiGyElLoWhAt

    User Avatar
    Gold Member

    Iw is moment of inertia * angular velocity
     
  9. Apr 10, 2014 #8
    what exactly is the angular velocity? Im not sure If I am up to that yet in my physics class. The section this problem was from was covering universal gravitation as well as kepler's laws. The question just says find the maximum and minimum velocities it doesn't specify between angular or tangential. The KE I am using wont work?
     
  10. Apr 10, 2014 #9

    BiGyElLoWhAt

    User Avatar
    Gold Member

    well angular velocity is the rate of revolution measured in radians/seconds, or also commonly 1/seconds (radians are funny that way). w = 2pi/T =v/r, so what you're doing will work I think.
     
  11. Apr 10, 2014 #10

    BiGyElLoWhAt

    User Avatar
    Gold Member

    just out of curiosity is it a general physic course?
     
  12. Apr 10, 2014 #11
    Suppose it was a perfectly circular orbit around the sun, at the average distance of 1.495x10^11 m. Would you know how to determine the tangential velocity then (say using F = ma)? What value would you get for the tangential velocity?

    Chet
     
  13. Apr 10, 2014 #12

    vela

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    Using ##V_f = \frac{2\pi R_f}{T}## isn't correct. This assumes that the Earth covers a distance of ##2\pi R_f## in one year, which would be true if it followed a circular path of radius ##R_f##, but the Earth covers less distance because it moves toward the Sun.

    You have two unknowns, so you're going to need another equation. You used conservation of energy already (though the sign of the potential energy is wrong). What's another conserved quantity you can use?
     
  14. Apr 10, 2014 #13

    vela

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    This is completely wrong. You've misread the problem.
     
  15. Apr 10, 2014 #14
    I think energy is the only conserved quantity I know so far. The only one that definitely stays the same over time. Also, do you mean another equation involving velocity?
     
  16. Apr 10, 2014 #15

    ehild

    User Avatar
    Homework Helper

    What about angular momentum?

    ehild
     
  17. Apr 10, 2014 #16

    vela

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    It's a central force, so angular momentum is conserved.
     
  18. Apr 11, 2014 #17
    I havent learned about momentum yet. There is no other way to solve this?
     
  19. Apr 11, 2014 #18

    vela

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    Use F=ma to get another relationship between V and R.
     
  20. Apr 11, 2014 #19

    ehild

    User Avatar
    Homework Helper

    You certainly know Kepler's Laws. What do the first two say?

    ehild
     
  21. Apr 11, 2014 #20
    do i have to use T^2 = r^3
     
  22. Apr 11, 2014 #21
    no wait thats the third. the first one says that the orbits of planets around the sun are ellipses. The second law says that each planets moves so that a line from the planet to the sun sweeps out equal areas in equal time
     
  23. Apr 11, 2014 #22

    ehild

    User Avatar
    Homework Helper

    That is it, the second one. What are the areas the line from the planet to the Sun sweeps in one second when the planet is nearest and when it is farthest? For that short time you can consider those parts of the ellipse as they were sectors of circles. How do you calculate the area of a circular sector?

    ehild
     
  24. Apr 11, 2014 #23
    the area would be (theta * r^2)/2 and since it is going from the farthest point to the nearest point its only sweeping out half the ellipse so theta would be pi?
     
  25. Apr 11, 2014 #24

    vela

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    I think using F=ma would be simpler.
     
  26. Apr 11, 2014 #25

    ehild

    User Avatar
    Homework Helper

    I mean only thin sectors, swept by the radius in 1 second at the nearest and farthest points.
    You know that the area of a circular sector is sr/2, where s is the length of arc and r is the radius of the circle. Supposing the speed of the planet is v, what is the length s it covers in 1 s?

    ehild
     

    Attached Files:

Share this great discussion with others via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted