Suppose a ball of finite mass is taken to such a height in space that the gravitational acceleration decreases significantly. Now, as you let go of the ball, it should head straight down towards the surface of earth, and as it crosses distance and gets closer to the earth surface, it's acceleration should increase. How would I, in this case, be able to calculate the distance traveled by the ball in a certain time?


$$\begin{align}F&=-\frac {GMm}{x^2}\\ a&=-\frac {GM}{x^2}\\ v\frac {dv}{dx}&=-\frac {GM}{x^2}\\ \int v\;{dv}&=\int-\frac {GM}{x^2}\;dx\\ \frac {v^2}2&=\frac {GM}x\\ v&=\sqrt\frac {2GM}x\\ \int\sqrt x\;dx&=\int\sqrt {2GM}\;dt\\ \frac 23x^{\frac 32}&=t\sqrt {2GM} \end{align}$$

Here, you have an equation relating distance and time in a gravitational field.

  • $\begingroup$ This is valid only when the object has zero total energy, which is not the case in the OP’s question. And, with these signs, the object is rising, not falling as the OP asked. $\endgroup$
    – G. Smith
    Feb 8 '20 at 17:25
  • $\begingroup$ I derived a general equation and as this is a homework question, I find it reasonable to let the OP figure out for himself how to use it. $\endgroup$
    – Sam
    Feb 8 '20 at 18:28
  • $\begingroup$ It is not a general equation at all. It applies to a specific case which is not the requested case. $\endgroup$
    – G. Smith
    Feb 8 '20 at 18:29
  • $\begingroup$ The general equation is in the Wikipedia link. There is no simple formula for position as a function of time, only for time as a function of position. There is no indication that this is a homework problem. In fact, it usually is not given as homework because the integral is messy. Your case is the simple homework case. $\endgroup$
    – G. Smith
    Feb 8 '20 at 18:50

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