Calculating Height of Thrown Object w/ Changing Deceleration g

[JohnnyGui]

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.

 

[jrmichler]

I might be missing something very obvious.

Then start with a sanity check:
1) Try a small velocity, say 100 m/sec, and compare to H = V^2/2g.
2) Try escape velocity for Earth, and see if you get infinite height.

 

[Ibix]

You know the acceleration due to gravity at radius $r$. Acceleration is $dv/dt$. So you have a differential equation in terms of v, t and r. It would be helpful if you could eliminate the dt somehow - then you could integrate to get $v(r)$ and insert boundary conditions.

Sanity check - your answer should be the same as your conservation of energy scheme.

 

[mfig]

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.

In this equation, t must be the time it takes the object to reach H, not an independent variable. If t is supposed to be a variable then the equation is simply wrong because you have a constant on the RHS and a quadratic in t on the LHS. You need to take into account that the time to reach the final height is itself a function of the gravitational force.

 

[JohnnyGui]

In this equation, t must be the time it takes the object to reach H, not an independent variable. If t is supposed to be a variable then the equation is simply wrong because you have a constant on the RHS and a quadratic in t on the LHS. You need to take into account that the time to reach the final height is itself a function of the gravitational force.

Yes, I agree with you. The problem is however that I can't write $t$ in terms of just $g$ since there are too many $t$'s in the formula.

@Ibix : Thanks. Do you mean by this that $\frac{GM}{r^2} = \frac{dv}{dt}$ and use this to eliminate the $t$'s in the formula?

@jrmichler: Just did. As I expected, a velocity of 100m/sec gave quite similar answers but using the escape velocity didn't give an infinite height with my mentioned formula since I'm using a fixed $g$ obviously.

 

[Ibix]

@Ibix : Thanks. Do you mean by this that $\frac{GM}{r^2} = \frac{dv}{dt}$ and use this to eliminate the $t$'s in the formula?

Not quite. The formula you originally developed doesn't help you because you've tried to insert the boundary conditions without solving the differential equation. You have now correctly written the differential equation (give or take a minus sign), but it's tricky to solve because you don't know $r (t)$. It would be easy, though, if the right hand side said something like $dv/dr$. Any ideas how to achieve that?

 

[Nugatory]

If you just need the height the object reaches, it is much much easier to use energy methods - you know the initial kinetic energy and it's easy to solve for the potential as a function of height. You only need to mess with that (not altogether trivial) differential equation if you want the trajectory as a function of time.

 

[JohnnyGui]

If you just need the height the object reaches, it is much much easier to use energy methods - you know the initial kinetic energy and it's easy to solve for the potential as a function of height. You only need to mess with that (not altogether trivial) differential equation if you want the trajectory as a function of time.

Yes, that's indeed what I want to know.

Not quite. The formula you originally developed doesn't help you because you've tried to insert the boundary conditions without solving the differential equation. You have now correctly written the differential equation (give or take a minus sign), but it's tricky to solve because you don't know $r (t)$. It would be easy, though, if the right hand side said something like $dv/dr$. Any ideas how to achieve that?

Thanks for the tip, I tried formulating $\frac{dv}{dr}$. Here it goes.
Let $dH$ be $dr$. I concluded that $g[r]$, which is the deceleration at height $r$, doesn't (merely) change until the object reaches height $r + dr$
Using this information, this means that:
$$dr = v[r] \cdot t -\frac{GM}{2r^2}\cdot t^2$$
Furthermore, the change in velocity per each $dr$ is equal to $dv=\frac{GM}{r^2} \cdot t$. Dividing $dv$ by $dr$ should give the differential equation of the function $v[r]$.
$$v'[r] = \frac{dv}{dr} = \frac{\frac{GM}{r^2}\cdot t}{v[r]\cdot t - \frac{GM}{2r^2}\cdot t^2} = \frac{1}{\frac{v[r]\cdot r^2}{GM}-\frac{t}{2}}$$
This means that I can get $t$ from:
$$t=2 \cdot (\frac{v[r] \cdot r^2}{GM} - \frac{1}{v'[r]})$$
Does this make sense so far?

 

[Ibix]

Using this information, this means that:
$$dr = v[r] \cdot t -\frac{GM}{2r^2}\cdot t^2$$

This can't be right - you've got an infinitesimal on one side and a finite quantity on the other. Either the left hand side has to be dr/dsomething or the right had side needs a dsomething.

In any case, you're working far harder than you need to. Are you familiar with the chain rule? Can you use it to express dv/dt in terms of dv/dr?

 

[JohnnyGui]

This can't be right - you've got an infinitesimal on one side and a finite quantity on the other. Either the left hand side has to be dr/dsomething or the right had side needs a dsomething.

You're right, I missed this. The dsomething is exactly what I'm struggling with since there are too many variables in the function that are also dependent on each other. Shouldn't the time $dt$ corresponding to a distance $dr$ be like:
$$dt=\frac{dr}{\frac{1}{2}(v[r] + v[r +dr])}$$
And:
$$\frac{1}{2}dv = \frac{1}{2} \cdot v'[r] \cdot dr$$
So if I multiply these two functions together, I should get $dr$? I'm deliberately multiplying the half of $dv$ with $dt$ for getting $dr$, since the deceleration $g$ is considered constant within a distance $dr$; in that case the distance $dr$ is therefore the mean of $dv$ multiplied by $dt$.

In any case, you're working far harder than you need to. Are you familiar with the chain rule? Can you use it to express dv/dt in terms of dv/dr?

I'll look into that and see what I can get out from it.

 

[JohnnyGui]

Sorry, but could someone please verify my mentioned formula's in my above post #10? I want to make sure if I understand these relationships correctly.

 

[Ibix]

I don't follow what you were trying to do. I was just going to write $$\frac{dv}{dt}=\frac{dv}{dr}\frac{dr}{dt}=\frac{dv}{dr}v$$That ought to be fairly straightforward to integrate.