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Trying to find the analytical equation for instantaneous velocity of an object under the influence of gravity (non-uniform acceleration) in a straight line. I am trying to model a stationary massive object (ex. sun) of mass M attracting a small ball of mass m and trying to find the small object's velocity and position. I know the force would be given by:

$$F = \frac{G M m}{d^2}$$

and that the acceleration of small object is $F/m$. I also know that the integral of acceleration with respect to time $t$ equals the velocity. So does that mean the equation for velocity would be:

$\large{v(t) = v_0 + \frac{GM}{d^2}t}$

and position would be

$\large{d(t) = d_0 + v_0t + \frac{1}{2}\frac{GM}{d^2}t}$

that doesn't seem right. I know since its non-uniform acceleration its not that simple but i don't know how to solve it. Any help would be greatly appreciated.

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  • $\begingroup$ This straight line motion is known as a radial trajectory, and the calculus for it isn't that hard to do, but it's still nice to see a worked solution. :) $\endgroup$
    – PM 2Ring
    Commented Jan 11, 2017 at 13:28
  • $\begingroup$ so is this solution right? $\endgroup$
    – user3712173
    Commented Jan 11, 2017 at 13:41
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    $\begingroup$ Go do some research on orbital mechanics/ Kepler problem. Every sophomore level mechanics book (even engineering books like Beer & Johnston) have solutions to this. Investigate Taylor's Classical Mechanics or Symon's Mechanics. $\endgroup$
    – Bill N
    Commented Jan 11, 2017 at 16:21
  • $\begingroup$ The acceleration is not constant. It changes with distance. The SUVAT equations only apply for constant acceleration. $\endgroup$
    – sammy gerbil
    Commented Jan 12, 2017 at 2:39
  • $\begingroup$ Related: physics.stackexchange.com/q/19813 and physics.stackexchange.com/q/63590 (probably among others) $\endgroup$
    – Kyle Kanos
    Commented Jan 12, 2017 at 11:18

1 Answer 1

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No, your result is incorrect. Remember that $d$ is a function of $t$, so $\frac{G M}{d^2}$ is not a constant and your result for $v(T)$ (and consequently $d(t)$, for the same reason) is wrong. You need to solve the differential equation $\ddot d(t)=-\frac{G M}{d(t)^2}$.

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