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Homework Help: Transforming Newton's Law of Universal Gravitation

  1. Jan 17, 2014 #1
    Hello, physicsforums. I'm trying to write a proof for a function involving Newton's law of gravitation, and I seem to be stuck. The function I'm trying to build is a function of time with respect to distance.

    This is the formula I want to transform.

    For Acceleration, I expanded it to

    [itex]\mathrm{A}=\frac{\mathrm{d}v}{\mathrm{d}x}\frac{\mathrm{d}x}{\mathrm{d }t}=\mathrm{V}\frac{\mathrm{d}v}{\mathrm{d}x}[/itex]

    and so the equation becomes


    To which I integrate and simplify to


    But V can be rewritten as dx/dt, so I've rewritten the problem to


    and integrate again to find time T.
    Now, this is where I think I have my problem, but I've checked my math and can't find anything wrong.
    It integrates and simplifies to


    However, when I plug the formula in with M = mass of the sun (1.989*10^30 kg) and x = distance between earth and the sun( 1.496*10^11 meters ), the formula tells me it would take about 27 days to fall into the sun, when I know it would really take about 64 days.

    I looked for someone who asked the same question at http://www.physlink.com/Education/AskExperts/ae226.cfm but do not understand how they got a different formula. I am trying to also stay away from Kepler's Laws, so this should be a pure integration/transformation problem.
  2. jcsd
  3. Jan 17, 2014 #2

    D H

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    There are two problems here. One is the sign. Your x is the distance from the Sun. Since the object is falling toward the Sun, your x(t) is a decreasing function of time. Thus v(t) is negative. The other problem is the constant of integration. Presumably the object starts at rest (v=0) at some initial distance x0 from the Sun.

    Correct this and you should (eventually) wind up with the correct result.
  4. Jan 17, 2014 #3


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    What happened to the constant of integration? What is the distance when velocity is zero?
  5. Jan 17, 2014 #4


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    x is decreasing and v is increasing, both are always positive, that is fine.

    The missing constant of integration is the problem. 27 days is the time an object, coming from infinity, needs for the distance earth<->sun.
  6. Jan 17, 2014 #5
    Well this is assuming that Vi = 0, so it's like a free-fall problem. This should make it much simpler.

    How did you get that number (27 days)? And how can I find the missing constant of integration?
    Can you please walk me through it?
  7. Jan 17, 2014 #6

    D H

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    Then his next line is wrong, where he uses v=dx/dt. One way or another he has a sign problem.

    Yep. The missing constant of integration is the big problem.
  8. Jan 17, 2014 #7
    How do I find it?
    I can't seem to work it through by myself.
  9. Jan 17, 2014 #8

    D H

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    The velocity is zero when you are a distance x0. That gives you the constant of integration.
  10. Jan 17, 2014 #9
    I know that the final answer after all the integration should be
    [itex]\mathrm{T}=\frac{\pi x^{3/2}}{2\sqrt{2GM}}[/itex]

    But I can't seem to get anywhere remotely close. Maybe I'm not getting the right constant? (I tried plugging it in at distance [itex]\mathrm{x}_0=1.496\times10^{11}[/itex] meters into [itex]\mathrm{0}m/s=-\sqrt{\frac{2\mathrm{GM}}{x}+\mathrm{C}}[/itex] so that [itex]\mathrm{C}=-5.63821\times 10^{-10}[/itex] when M=mass of the sun (1.989*10^30 kg ) and G being the Gravitational Constant. I think I'm going in the wrong direction but I don't know where. At this point, I feel like I"m walking blindly. Can somebody build this proof and show me how they got there? I think I'm missing something very important (or made some very stupid error)
    Last edited: Jan 17, 2014
  11. Jan 17, 2014 #10

    D H

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    Better, but don't be so quick to go to numbers. Keep things symbolic as long as possible.

    Don't use G*Msun. Use the solar gravitational parameter μsun, which is conceptually short for G*Msun, but is (a) a single symbol rather than two, and (b) a whole lot more precise than G*Msun. G is only known to four places or so, while astronomers know μsun to ten places or so.

    Keeping things symbolic, your velocity is ##v^2 = 2\mu\left(\frac 1 x - \frac 1 {x_0}\right)##, where x0 is the distance at which the object is released with zero velocity.
  12. Jan 17, 2014 #11


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    For maximum accuracy, compare the formula with that for Earth's period of orbit. The two should differ only by a factor consisting of mathematical constants.
  13. Jan 17, 2014 #12
    This is correct for the final answer. [itex]x[/itex] is the distance to Earth, plugging in the appropriate numbers returns 64 days. (To the amount of sigfigs I was using, it was calculated at about 64.7 days)

    DH has good information as to realize what your constant should be. Attention should be put into realizing your limits of integration. In fact, in the above equation you should get [itex]x=x_{0}[/itex], define as in DH post.
  14. Jan 17, 2014 #13
    What about the constant of integration (C)?
    I'm confused now. You told me earlier that velocity is zero when at distance x0. How am I going to find the constant of integration if I keep everything symbolic? Or is x0 the important constant you're referring to?

    Also, when I integrate [itex]\int\mathrm{d}t=\int\frac{\mathrm{d}x}{\sqrt{2\mu(\frac{1}{x}-\frac{1}{x_0})}}[/itex], would that mean that 'x' be my variable of integration and 'x0' be treated like a constant? And if so, why is it in this way?
  15. Jan 17, 2014 #14

    D H

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    For maximum accuracy you should use μsun=132,712,440,018±9 km3/s2. This value is based not just on the Earth's orbit but on observations of the Earth's orbit, observations of all of the planets, and observations of eclipses, some of which go back thousands of years.

    x0 essentially is the constant of integration. One key difference between your 5.6×10-10 and my 2μ/x0 is that I am keeping the constant of integration symbolic. Another is that I wrote your C as 2μ/x0. They are the same number.

    Here's another way to arrive at this result: Look at the problem as a physics problem first, a math problem second. You did it the other way around. Gravitation is a conservative force. This means the total mechanical energy of the falling object is conserved:
    [tex]\frac 1 2 v^2 - \frac {\mu} x = \text{constant}[/tex]
    That constant is easily evaluated since it is a given that the velocity is zero at x=x0. Thus
    [tex]\frac 1 2 v^2 - \frac {\mu} x = -\,\frac {\mu} x_0[/tex]
    [tex]v^2 = 2\mu \left(\frac 1 x - \frac 1 {x_0}\right)[/tex]

    Of course.

    Because x is a variable. It's your variable. Depending on how you look at it, x0 is a constant of integration (so it's a constant) or it's an arbitrary point where you fix the potential (so it's a constant).
  16. Jan 17, 2014 #15
    Sorry, I am starting to feel like I am interjecting. Lets look at it one more way, when integrating your equation you are actually using definite integrals, so the constant is defined as explained above. Here is an example, without being too explicit

    [itex]\int_i^f dv = \int_i^f dx[/itex],

    where [itex]i[/itex] and [itex]f[/itex] are initial and final, respectfully. In your problem the initial velocity was zero and the final velocity was [itex]V[/itex]. When your velocity is 0 is when [itex]x=x_0[/itex] and you integrate to an arbitrary x. What is [itex]x_0[/itex]? Think about it.

    This yielded your [itex]V[/itex], which actually has two solution of + and -. In this case the velocity is traveling inward towards the origin of your system and therefore should be a negative. Otherwise there is a sign problem going into the final result.

    Your final integral has to be taken in a similar manner. Though this time both limits are well known to you.

    Ok, last post. I was just trying to add a different perspective. Once again sorry, your already doing a good job DH.
  17. Jan 17, 2014 #16
    Okay, I think I'm starting to get it. Thank you for the help, DH, for making that clear connection between the integral constant C and x0. I'm making an important mental note not to change the form of definite integrals to simplify constants, as I may need them for later integrations.

    From here, I've worked out that I'm supposed to integrate like so, to find time. [itex]\sqrt{2\mu}\int\mathrm{d}t=-\int\frac{\mathrm{d}x}{\sqrt{\frac{1}{x}-\frac{1}{x_0}}}[/itex]

    And this makes perfect sense, and I can also see the [itex]\int ... = \mathrm{T}\sqrt{2\mu}[/itex] within phsyslink I posted earlier.
    This integral seems tough to solve though. phsyclink implies it resolves to an inverse trig function, but wolfram says here that an integral in that form resolves something else entirely (involving a natural logarithm function?). What do I make of this?
  18. Jan 17, 2014 #17

    D H

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    Mathematica (the engine behind Wolfram Alpha) *loves* to jump to complex solutions. Sometimes it does so even when you say not to.

    Try a trig substitution. This isn't at all obvious, so I'll give away the farm. Try ##\cos^2\theta = \frac x{x_0}##.
  19. Jan 18, 2014 #18

    D H

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    I'm going to take this as a teaching moment on the perils of symbolic mathematics software such as Mathematica. The advantages are obvious. The disadvantages, not so. This is one of them.

    Mathematica (and Maple, Maxima, etc.) are wont to use complex algebra for two key reasons: The fundamental theorem of algebra and the niceties of holomorphic functions. It's the easy way out. Sometimes that easy way out hides the solution you want. That definitely is the case with this integral. The solution proffered by WA includes the following:
    [tex]\frac 1 2 i c^{\,3/2} \log\left(2\sqrt{cx(c-x)} - i\sqrt{c}(c-2x)\right) + \text{constant}[/tex]
    The argument to the natural log is ##2\sqrt{cx(c-x)} - i\sqrt{c}(c-2x)##. If ##c \ge x \ge 0##, this is a complex number with magnitude ##c^{3/2}## and some argument ##\theta##. Thus this term can be expressed as ##\log(c^{3/2}\exp(i\theta))##. The product ##c^{3/2}\exp(i\theta)## becomes addition after taking the logarithm. The ##\log(c^{3/2})## term can be tossed as just another part of the constant of integration. What's left after taking the log and multiplying by ##i c^{3/2}/2## is just ##-c^{3/2}\theta/2##. That theta? That's a trig substitution, hidden in that complex logarithm mess.
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