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You probably know the the pull of gravity varies with the inverse square of the distance $(r = x)$ $x^2$. We all know in physics that the integral of $\frac{K}{x^2}$ is $-\frac{K}{x}$, this when the attracted body is free-falling.

I'd like to ask for your help with a particular case when the body is travelling at great speed, at say $.99c$ : it will reach B in only $100$ seconds and will therefore absorb a smaller amount of energy.

I am not a great expert of maths, but I have a hunch I should get the exact value dividing available energy by speed,so, the energy absorbed should be $$\frac{9.9\times10^{13}}{10^{10}} = 9900$$

I tried to sum up the $100$ individual values I got from $100$ divisions, but I get no more than $7000$. Can you tell me how tho find the correct integral?

I draw this sketch if it helps

Sketch

user157860
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    To answer this question in the relativistic framework, you need to be more rigorous, rewrite $f=ma$ with the appropriate corrections and integrate the differential equation. –  Jul 25 '18 at 09:05
  • I want to mention two things. First of all try touse MathJax (https://math.meta.stackexchange.com/questions/5020/mathjax-basic-tutorial-and-quick-reference) the next time you are asking a question. The second one is about the integral of $\int \frac{K}{x^2} dx = \frac{K}{x}$. To be exact it should be $\int \frac{K}{x^2} dx = -\frac{K}{x}$ or are you assuming that the minus sign is within the second constant? – mrtaurho Jul 25 '18 at 09:06
  • @mrtaurho,thanks, I edited and will try mathJax – user157860 Jul 25 '18 at 09:09
  • @YvesDaoust,thanks, I got the result with relativity at PSE, yet I'd like to know how to find such an integral, since I need it for other purposes.Is it correct to divide by the speed? – user157860 Jul 25 '18 at 09:12
  • What is PSE ? Can you show the relativistic equation ? –  Jul 25 '18 at 09:19
  • @YvesDaoust, probably they are OT here, you can find them at PSE – user157860 Jul 25 '18 at 10:01
  • This is quite insufficient, there are only equations for a constant acceleration. You need the relativistic gravitational field. –  Jul 25 '18 at 10:05
  • @YvesDaoust, what I am looking for here is just the correct sum of a great number of divisions by x^2, is it more than 7000? – user157860 Jul 25 '18 at 10:08
  • The sketch and the problem is not very clearly presented so I have no idea what you have done. But if you are trying to estimate an integral as a sum then this is done (schematically) as $\int_a^b f(x){\rm d}x \approx \sum f(x_i)\Delta x_i$. Did you just sum the values and forgot to multiply by the step lengths $\Delta x_i$? – Winther Jul 25 '18 at 15:38
  • "I have a hunch I should get the exact value dividing available energy by speed" Why? The energy you extract from the gravitational field by going from far away towards a massive body only depends on the displacement, not the speed you start off with. – Winther Jul 25 '18 at 15:42
  • @Winther,sorry I was not clear, I made the interval= 10^10, because in 1 second (the time necessary to absorb gravity) it travels that far. If you look at the bottom, there are 100 intervals of 10^10 cm. Then I summed up the 99 values at each interval. My hunch is suppported that the sum gives me 7000 and it is surely less than the exact value. It boils down to the integral of 1000/x^2 from 1 to 100, which is exactly 9900- Thanks for your help – user157860 Jul 25 '18 at 16:02
  • Ok. My bet is that 100 points is not enough to compute the integral accurately. Try with 1000 points. – Winther Jul 25 '18 at 16:39

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