To use a slightly different problem pair, because it would be easier for me to compute:
Problem one. I have mass of 80kg at point [10,0] (simplifying to two dimensions, as I don’t need Z). A 2,000 KG object is resting at position [0, 0]. The Newtonian force of 1.0687 10^-7 N towards the origin should be accurate. [Edit: 1.06710^-6 N, when I calculated it again. Forgot to update this section]
Problem two. I have mass of 80kg at point [10,0] A 2,000 KG object is moving at 10 m/s along the Y axis, position defined as r(t) = [0, −50 + 10t]. Using strictly the time interval t = 0 → t = 10, where t is in seconds, calculating the force when t=5...
distance(t) = sqrt(10^2 + c^2((5 - t)^2)
Gravity(t) = 6.67 10^-11 sum(802,000distance(t), for t > 0, t < 10) (10 / distance(t)) [Strictly speaking, this should be an integral over the whole of t, not a summation on a limited subset of t, but I’m doing this the faster, slightly less accurate way; the 10 / distance(t) at the end is to take only the y portion of my vectors, as the t portion of the gravitational vectors cancel out.]
Which gives, not entirely surprisingly, 1.067 * 10^-6 N directed to the origin. (I think your calculation was off by an order of magnitude, I’m not sure why.)
The difference between Newtonian gravity and gravity with respect to y is 3.38 * 10 ^-33. Which is expected; if the difference in gravitational force were greater, it would have been noticed a long time ago.
I probably messed up somewhere in there, because my brain is mush and it’s been a while since I’ve mucked about with vectors, but this should give you the basic idea.
I must apologize for the delay in replying. Regretfully, I don’t think I can spare any more time for this exchange (and am going to be taking a break from this and some other addicting sites), so this will likely be my final reply.
distance(t) = sqrt(10^2 + c^2((5 - t)^2) Gravity(t) = 6.67 10^-11 sum(802,000distance(t), for t > 0, t < 10) * (10 / distance(t))
Now I think I sort-of see what you’re trying to do here, but I don’t understand what’s motivating that specific expression; it seems to me that if you want to treat space and time symmetrically, then the expression you want is something more like
(80)\,dt}{10%5E2+(-50+10t)%5E2+c%5E2(5%20-%20t)%5E2}), which should be able to be evaluated with the help of a standard integral table.
Please don’t interpret this as hostility (for this is the sort of forum where it’s actually considered polite to tell people this sort of thing), but my subjective impression is that you are confused in some way; I don’t have the time or physics expertise to fully examine all the ideas you’ve mentioned and explain in detail exactly why they fail or why they work, but what you’ve said so far has not impressed me. If you want to learn more about physics, you are of course aware that there are a wide variety of standard textbooks to choose from, and I wish you the best of luck in your future endeavors.
I do not interpret it, or any of your other responses, as hostility. (I’ve been upvoting your responses. I requested feedback, and you’ve provided it.)
I did indicate the integral would be more accurate; I can run a summation in a few seconds, however, where an integral requires breaking out a pencil and paper and skills I haven’t touched since college. It was a rough estimate, which I used strictly to show what it was such a test should be looking for. Since we aren’t running the test itself, accuracy didn’t seem particularly important, since the purpose was strictly demonstrative.
(Neither formula is actually correct for the idea, however. The constant would be be wrong, and would need to be adjusted so the gravitational force would be equivalent to the existing formula for an object at rest.)
To use a slightly different problem pair, because it would be easier for me to compute:
Problem one. I have mass of 80kg at point [10,0] (simplifying to two dimensions, as I don’t need Z). A 2,000 KG object is resting at position [0, 0]. The Newtonian force of 1.0687 10^-7 N towards the origin should be accurate. [Edit: 1.06710^-6 N, when I calculated it again. Forgot to update this section]
Problem two. I have mass of 80kg at point [10,0] A 2,000 KG object is moving at 10 m/s along the Y axis, position defined as r(t) = [0, −50 + 10t]. Using strictly the time interval t = 0 → t = 10, where t is in seconds, calculating the force when t=5...
distance(t) = sqrt(10^2 + c^2((5 - t)^2) Gravity(t) = 6.67 10^-11 sum(802,000distance(t), for t > 0, t < 10) (10 / distance(t)) [Strictly speaking, this should be an integral over the whole of t, not a summation on a limited subset of t, but I’m doing this the faster, slightly less accurate way; the 10 / distance(t) at the end is to take only the y portion of my vectors, as the t portion of the gravitational vectors cancel out.]
Which gives, not entirely surprisingly, 1.067 * 10^-6 N directed to the origin. (I think your calculation was off by an order of magnitude, I’m not sure why.)
The difference between Newtonian gravity and gravity with respect to y is 3.38 * 10 ^-33. Which is expected; if the difference in gravitational force were greater, it would have been noticed a long time ago.
I probably messed up somewhere in there, because my brain is mush and it’s been a while since I’ve mucked about with vectors, but this should give you the basic idea.
I must apologize for the delay in replying. Regretfully, I don’t think I can spare any more time for this exchange (and am going to be taking a break from this and some other addicting sites), so this will likely be my final reply.
Now I think I sort-of see what you’re trying to do here, but I don’t understand what’s motivating that specific expression; it seems to me that if you want to treat space and time symmetrically, then the expression you want is something more like
(80)\,dt}{10%5E2+(-50+10t)%5E2+c%5E2(5%20-%20t)%5E2}), which should be able to be evaluated with the help of a standard integral table.Please don’t interpret this as hostility (for this is the sort of forum where it’s actually considered polite to tell people this sort of thing), but my subjective impression is that you are confused in some way; I don’t have the time or physics expertise to fully examine all the ideas you’ve mentioned and explain in detail exactly why they fail or why they work, but what you’ve said so far has not impressed me. If you want to learn more about physics, you are of course aware that there are a wide variety of standard textbooks to choose from, and I wish you the best of luck in your future endeavors.
I do not interpret it, or any of your other responses, as hostility. (I’ve been upvoting your responses. I requested feedback, and you’ve provided it.)
I did indicate the integral would be more accurate; I can run a summation in a few seconds, however, where an integral requires breaking out a pencil and paper and skills I haven’t touched since college. It was a rough estimate, which I used strictly to show what it was such a test should be looking for. Since we aren’t running the test itself, accuracy didn’t seem particularly important, since the purpose was strictly demonstrative.
(Neither formula is actually correct for the idea, however. The constant would be be wrong, and would need to be adjusted so the gravitational force would be equivalent to the existing formula for an object at rest.)
Thank you for your time!