- #1
JohnnyGui
- 796
- 51
Good day,
I have a question regarding calculating the total reached height ##H## of an object that is thrown upwards with an initial velocity ##v_i##, while taking into account the decreasing gravitational deceleration ##g## with height as well.The formula would be something like:
$$v_i \cdot t - \frac{GM}{r^2} \cdot \frac{1}{2} \cdot t^2 = H$$
My problem here is that the deceleration (##\frac{GM}{r^2}##) inside the formula changes with the outcome of the formula itself, since ##H## represents the height.
My temptation is to integrate this formula but I’d have to integrate over the outcome ##dH## and not over the time ##dt##. Besides, integrating over the time ##dt## instead is risky for inaccuracy because the initial velocity might be very high and thus the object can reach in ##dt## time a very high distance in which the deceleration ##g## might change significantly.
Writing ##t## in terms of ##dH## is also not possible since it has two possible solutions for ##t## (one on the way up and one when the object is falling down).
Question: Is it actually possible to calculate the reached height ##H## with this formula while taking into account the changing deceleration ##g## over the height itself? I was able to deduce that there is another way of doing this by solving ##\frac{1}{2} m v_i^2 = GMm (\frac{1}{r} - \frac{1}{r + H})## but I was specifically wondering about my mentioned formula. I might be missing something very obvious.
I have a question regarding calculating the total reached height ##H## of an object that is thrown upwards with an initial velocity ##v_i##, while taking into account the decreasing gravitational deceleration ##g## with height as well.The formula would be something like:
$$v_i \cdot t - \frac{GM}{r^2} \cdot \frac{1}{2} \cdot t^2 = H$$
My problem here is that the deceleration (##\frac{GM}{r^2}##) inside the formula changes with the outcome of the formula itself, since ##H## represents the height.
My temptation is to integrate this formula but I’d have to integrate over the outcome ##dH## and not over the time ##dt##. Besides, integrating over the time ##dt## instead is risky for inaccuracy because the initial velocity might be very high and thus the object can reach in ##dt## time a very high distance in which the deceleration ##g## might change significantly.
Writing ##t## in terms of ##dH## is also not possible since it has two possible solutions for ##t## (one on the way up and one when the object is falling down).
Question: Is it actually possible to calculate the reached height ##H## with this formula while taking into account the changing deceleration ##g## over the height itself? I was able to deduce that there is another way of doing this by solving ##\frac{1}{2} m v_i^2 = GMm (\frac{1}{r} - \frac{1}{r + H})## but I was specifically wondering about my mentioned formula. I might be missing something very obvious.