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!

Gravitational forces, speeds and masses

  1. Apr 14, 2017 #1
    1. The problem statement, all variables and given/known data

    A comet moves around a stat in ##xy## plane along elliptical orbit, described by

    $$
    0.16 x^2 + y^2 = 4
    $$

    where ##x, y## are in ##AU##

    1) Sketch the comet in the ##x,y## coordinate system denoting all orbit parameters

    2) Find the semi major and minor axes of the orbit in ##AU##

    3) Assuming the star is located at ##x < 0##, find the eccentricity of the orbit
    and the ##x, y## coordinates of the star

    4) If the orbital speed of this comet at the perihelion is ##100 ## km/s, find
    its orbital speed at aphelion

    5) Using the energy conservation approach find the approximate mass of the star
    to an order of magnitude.


    2. Relevant equations


    Eccentricity ##e = \sqrt{1 - \frac{b^2}{a^2}}##


    f ## = \sqrt{a^2 - b^2}## , I'm not sure what this is other than the distance
    between where the semi-major and minor axis meet and the location of the star.

    3. The attempt at a solution


    For part 1) I have drawn a sketch.

    For part 2) I note that the given expression is


    $$
    0.16 x^2 + y^2 = 4
    $$

    Plugging ##x = 0## into this gives ##y^2 = 4## so ##y = 2##. Then setting ##y = 0##
    gives ##x^2 = \frac{4}{0.16}## so ##x = 5##.

    This gives us our semi-major axis at ##5## and the semi-minor at ##2##, both in ##AU##.

    Part 3) Using the formula for eccentricity as

    $$
    e = \sqrt{1 - \frac{4}{25}} = \frac{\sqrt{21}}{5} \approx 0.917
    $$

    The ##x, y## coordinates of the star are found using ##f = \sqrt{a^2 - b^2}## which
    is

    $$
    f = \sqrt{25 - 4} = \sqrt{21} \approx 4.583
    $$

    From this we note that the location of the star is ##(-4.583, 0)## to 3 decimal
    places.

    part 4)

    I know that the perihelion has the fastest speed - but I'm not sure how to go
    about finding the speed of the aphelion given this information.

    Should I be setting up some kind of differential equation? Or is there an
    appropriate formula?

    part 5)

    I'm not sure how to consider this either. If the energy is conserved then I have
    the definition

    $$
    W + PE_0 + KE_0 =
    PE_f + KE_f + \text{Energy(Lost)}
    $$

    Given that there's no work being done in the motion and there's no energy lost
    due to friction we have

    $$
    PE_0 + KE_0 =
    PE_f + KE_f
    $$

    I'm not sure what to do with this though, in terms of finding the mass.

    There's no 'at rest' for this system is there, in the way that this expression
    is often used.

    The potential energy would be a result of the gravitational field, which is

    $$
    \text{Gravitational force } = \frac{-GMm}{x}
    $$

    Where ##x## is the distance between the orbiting body and the star, ##G## is the
    gravitational constant, ##M## is the mass of the star (what we want) and ##m## is
    the mass of the orbiting body (which we don't have either).


    Thanks
     
  2. jcsd
  3. Apr 14, 2017 #2

    gneill

    User Avatar

    Staff: Mentor

    For part (4) consider angular momentum.
     
  4. Apr 15, 2017 #3
    cheers - so for part 4 if I consider the angular momentum ##\omega## then:

    ##v = \omega r##, and here we have ##r = a##, the semi-major axis and the speed is
    ##100##km/s.

    So we have

    $$
    \frac{v}{r} = \omega
    $$

    Which gives us ##\omega = \frac{100}{5} = 20##.

    But this doesn't take into consideration the position of the star?
    I've not used any information about the mass, star position, just treated it
    like it was a uniform circular motion (i think?)
     
  5. Apr 15, 2017 #4

    gneill

    User Avatar

    Staff: Mentor

    Right. The motion is about the star and the velocity is only perpendicular to the radius at the ends of the major axis (The scalar version of ##\vec{h} = \vec{r} \times \vec{v}##, namely h = rv, needs r and v to be at right angles just as in circular motion). That means you need to use the perihelion and aphelion distances, so they want you to calculate those first (Don't worry, you'll need them for part 5 as well ).
     
  6. Apr 15, 2017 #5
    OK - so I have that ##f =\sqrt{21}##, and this is the distance from the star to
    the 'origin' (where major / minor meet). So from this I can say that the
    distance from the star to the orbiting body is

    $$
    dist = major - f
    $$

    Which here is

    $$
    dist = 5 - \sqrt{21} = 0.417242305
    $$

    So that I have the perihelion distance as ##\approx 0.4172##

    The aphelion distance would be (I think)

    $$
    2 \times \text{semi major} - (\text{perihelion distance})
    $$

    Which here is

    $$
    10 - 0.4172 = 5 + \sqrt{21}
    $$

    So I have these two distances, and need to use the angular momentum to find the
    mass of the star.

    I'm still considering angular momentum to be ##\omega## which is defined as
    ##\omega = \frac{\theta}{T}## where ##\frac{1}{t} = f = ## frequency.

    But from this - I'm not sure what the frequency really is?

    Perhaps If I use

    $$
    v = \omega r
    $$

    Then I can sub ##r## for the perihelion distance and express this as

    $$
    100 = \omega \times (\text{perihelion distance}) = \omega \times (5 - \sqrt{21})
    $$


    From this I have that

    $$
    \frac{100}{5 - \sqrt{21}} = \omega \approx 239.5643924
    $$

    So this is my angular momentum , which I may presume is a constant.

    Then If I have this as ##\omega## I can use it to find speed as

    $$
    v = \omega \times r
    $$

    Then

    $$
    v = 239.5643924 \times (5 - \sqrt{21}) \approx 2285.64
    $$

    Ouch, this is way off, as the speed must be slower here.

    I'm confused.
     
  7. Apr 15, 2017 #6

    gneill

    User Avatar

    Staff: Mentor

    Okay, your two distances look good.
    ##\omega## is angular velocity. For a point mass, angular momentum is given by ##\vec{L} = m \vec{r} \times \vec{v}##.

    Usually for orbital mechanics we consider "specific" quantities (per unit mass) since we often don't know or care about the mass of the orbiting object as it's usually insignificant with respect to the mass of the primary. So, dropping the "m" the specific angular momentum is given by

    ##\vec{h} = \vec{r} \times \vec{v}##

    When ##\vec{r}## and ##\vec{v}## are perpendicular (at the ends of the major axis), you have the scalar relationship ##h = r v##.
     
  8. Apr 15, 2017 #7
    Thanks - are you just using ##\vec{h}## here to represent angular momentum for
    this case? (I've not seen ##\vec{h}##, just making sure that's what this was for
    here).

    Then from this we have that

    $$
    h = rv
    $$

    So for this problem we have, for the perihelion position (where the object is closest)

    $$
    h = 100 \times (5 - \sqrt{21}) \approx 41.7424
    $$

    So then at the other end I have that the speed is

    $$
    41.7424 = v \times (5 + \sqrt{21})
    $$

    Then

    $$
    v = \frac{41.7424}{5 + \sqrt{21}} \approx 4.35607
    $$


    Therefore the speed at the aphelion position would be ##v = 4.35607##.

    For part 5 then I need to consider the mass of the object.

    I'm not sure what equations I should use for this.
     
  9. Apr 15, 2017 #8

    gneill

    User Avatar

    Staff: Mentor

    Yes. h is the commonly used variable for the specific angular momentum in orbital dynamics.
    Good. Be sure to tack on the appropriate units and round to an appropriate number of sig figs.
    The mass of the object will drop out along the way if you choose to include it in the workings. The problem gave you the hint of using an energy conservation approach. What energy conservation formula applies to orbits?
     
  10. Apr 15, 2017 #9
    So we have

    $$
    \frac{1}{2}mv_1^2 - \frac{GMm}{r_1} = \frac{1}{2}mv_2^2 - \frac{GMm}{r_2}
    $$



    Here we want to solve for ##M##, and can see that every term has ##m## in it (nice!)
    so can just drop that :)

    Dropping ##m## and multiplying through by 2 gives

    $$
    v_1^2 - \frac{2GM}{r_1} = v_2^2 - \frac{2GM}{r_2}
    $$

    Then

    $$
    v_1^2 - v_2^2 =
    \frac{2GM}{r_1}
    - \frac{2GM}{r_2}
    $$

    So

    $$
    M = \frac{v_1^2 - v_2^2}{\frac{2G}{r_1} - \frac{2G}{r_2}}
    $$

    And from here it's a case of plugging in the known values.

    I'm not too sure which ##r## is which here though.
    I think that I would have

    ##r_1## as the perihelion distance and ##r_2## as the aphelion distance (from the star).
     
  11. Apr 15, 2017 #10

    gneill

    User Avatar

    Staff: Mentor

    Yup.
     
  12. Apr 15, 2017 #11
    Thanks - so that's this question done then?
     
  13. Apr 15, 2017 #12

    gneill

    User Avatar

    Staff: Mentor

    Looks like it to me. What final value did you obtain for the mass of the star?
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Gravitational forces, speeds and masses
  1. Mass, force, speed? (Replies: 2)

Loading...