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I've been trying to solve this by myself without success. I am trying to find the velocity of a falling object as its height varies (change in velocity as it falls). Although it is often assumed that the acceleration is constant, I know that it isn't due to Newton's Law of Gravitation, where a higher height will cause acceleration to be lower. So I started off with this.

$$a=G\frac{M}{\left(R+h\right)^2}$$

Then, I got a relation with velocity:

$$a=\frac{\text{d}v}{\text{d}t}=\frac{\text{d}v}{\text{d}h}\times\frac{\text{d}h}{\text{d}t}=v\times\frac{\text{d}v}{\text{d}h}$$

So to find velocity in terms of h, I integrated both sides w.r.t. h and came to $$\int_{}^{}a(h)dh=\frac{1}{2}v(h)^2+C$$

But I didn't know what to do with that constant and after doing some research I realised that I had to make it in a definite integral form, with set values. So I had final position $h_f=0$ since that would be when the object comes to a stop (hits the ground) with $h$ as a changing variable and initial velocity $v_0=0$ since at release, the velocity is 0. So I tried: $$\int_{h}^{0}a(h)dh=\frac 12\left(v_0^2-v(h)^2\right)=-\frac{1}{2}v(h)^2$$

I integrated $a$ from above: $$\int_{h}^{0}\frac{GM}{(R+h)^2}dh=\left[-\frac{GM}{R+h}\right]_h^0=-\frac{GM}{R}+\frac{GM}{R+h}$$

Putting this with the equation of velocity above gave:

$$v(h)=±\sqrt{2\left(\frac{GM}{R}-\frac{GM}{R+h}\right)}$$

I assumed that the velocity would be negative since the object is falling. But when I tried to graph this, I realised that as $h$ gets smaller, or as the object falls, the velocity also decreases. This doesn't really make sense to me because wouldn't the velocity increase as an object keeps falling, meaning that as $h$ decreases, the velocity should increase? Does this velocity represent something else? Or did I simply do something wrong?

Also, how would I get to obtaining a height or velocity function against time?


EDIT: Okay, so I've tried solving this with an initial height set in mind and adding a negative in front of the acceleration. So, $h_0=50\times10^3$m and I put the final height as an unknown (negative values would be ignored). Doing this allowed me to $$\int_{50\times10^3}^{h}-\frac{GM}{(R+h)^2}dh=\left[\frac{GM}{R+h}\right]_{50\times10^3}^h=-\frac{GM}{R+h}+\frac{GM}{R+50\times10^3} =\left[\frac{1}{2}v(h)^2\right]_{50\times10^3}^h=\frac{1}{2}v(h)^2$$

This gave me a new equation of, $$v(h)=\sqrt{2\left(\frac{GM}{R+h}-\frac{GM}{R+50\times10^3}\right)}$$

This new equation starts off at 0 velocity when $h=50\times10^3$ and increases as $h$ decreases. This seems more of a reasonable result to me. Does this mean that my equation needs to have another variable, $h_0$, that also affects the velocity?

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    $\begingroup$ You did everything right, but the gravitational acceleration has a minus sign there. $\endgroup$ – J. Murray Oct 5 '17 at 15:27
  • $\begingroup$ It is far easier to use the conservation of energy. $\endgroup$ – sammy gerbil Oct 7 '17 at 9:17
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From energy conservation

$$\dfrac{mv^2}{2}=\frac{GMm}{R+h}-\frac{GMm}{R+h_o}$$

We immediately obtain your solution

$$v=\sqrt{2GM\left(\frac{1}{R+h}-\frac{1}{R+h_o}\right)}\tag{A}$$

Or

$$-\frac{dh}{dt}=\sqrt{\frac{2GM}{R+h_o}}\cdot\sqrt{\dfrac{h_0-h}{R+h}}$$

By integrating

$$-\int_{h_o}^h\sqrt{\frac{R+h}{h_0-h}}\cdot dh=\sqrt{\frac{2GM}{R+h_o}}\space \int_{t_o}^t dt$$

We obtain the exact (but not explicit) solution for $h(t)$

$$(R+h_0)\arctan\left(\sqrt{\dfrac{h_0-h}{R+h}}\right)+\sqrt{(h_0-h)(R+h)}=\sqrt{\frac{2GM}{R+h_o}}\space(t-t_o)\tag{B}$$

For a verification, simplify for $\space h\ll R$

Remember $\space\arctan(x)\approx x\space$ for $\space x\ll1$

$$\left(R+h_0\right)\sqrt{\dfrac{h_0-h}{R+h}}+\sqrt{(h_0-h)(R+h)}=\sqrt{\frac{2GM}{R+h_o}}\left(t-t_o\right)$$

And also $\space R+h\approx R\space$ for $\space h\ll R\space$ and $\space R+h_o\approx R\space$ for $\space h_o\ll R$

$$R\space\sqrt{\dfrac{h_0-h}{R}}+\sqrt{(h_0-h)R}=\sqrt{\frac{2GM}{R}}\left(t-t_o\right)$$

We obtain

$$\sqrt{(h_0-h)}=\sqrt{\frac{GM}{2R^2}}(t-t_o)$$

Or

$$h_0-h=\frac{g\space(t-t_o)^2}{2}$$

This is a well known formula where $g$ is the gravitational acceleration

$$\space g=\dfrac{GM}{R^2}$$

The solution for $v(t)$ consists of $(A)$ and $(B)$. It is not explicit, but is exact and can be solved numerically to any degree of precision.


EDIT: Integration steps

First, lets derive a specific reduction rule for integration. Note that

$$\int\dfrac{dx}{x^2+1}=\arctan{(x)}+C_1=-arctan{\frac{1}{x}}+C_2\tag{1}$$

Take a derivative

$$\dfrac{d}{dx}\left(\frac{x}{x^2+1}\right)=\frac{1}{x^2+1}-\frac{2x^2}{(x^2+1)^2}$$

And integrate it

$$\dfrac{x}{x^2+1}=\int\frac{dx}{x^2+1}-\int\frac{2x^2dx}{(x^2+1)^2}+C^{'}$$

By rearranging and using $(1)$, we obtain the reduction rule

$$-\int\dfrac{2x^2dx}{(x^2+1)^2}=\arctan\dfrac{1}{x}+\frac{x}{x^2+1}+C\tag{2}$$

Now define

$$x=\sqrt{\dfrac{R+h}{h_o-h}}\tag{3}$$

Solve for h

$$h_o-h=\dfrac{R+h_o}{x^2+1}\tag{4}$$

Differentiate (3)

$$\dfrac{dx}{dh}=\frac{1}{2x}\left(\frac{R+h}{(h_o-h)^2}+\frac{1}{h_o-h}\right)$$

Combine with (3) and (4)

$$\dfrac{dx}{dh}=\frac{1}{2x}\cdot\frac{x^2+1}{R+h_o}(x^2+1)$$

Rearrange

$$dh=(R+h_o)\dfrac{2xdx}{(x^2+1)^2}\tag{5}$$

From (3) and (5)

$$\int\sqrt{\dfrac{R+h}{h_o-h}}dh=\int xdh=(R+h_o)\int\dfrac{2x^2dx}{(x^2+1)^2}$$

By using (2) we obtain

$$-\int\sqrt{\dfrac{R+h}{h_o-h}}dh=(R+h_o)\left(\arctan\dfrac{1}{x}+\frac{x}{x^2+1}+C\right)$$

Finally by using (3) we obtain the solution (B). The lower integration limit $h=h_o$ defines $C=0$.

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  • $\begingroup$ I understand what you put up as the solution. I currently have another question (mostly about integration). Even though velocity is affected by initial h0 and the variable h, the h0 would act as a constant when integrating since h0 is determined by neither t nor h right? $\endgroup$ – Dan Skyler Oct 8 '17 at 16:20
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    $\begingroup$ Yes, this is correct, $h_o$ is a constant for integration. $\endgroup$ – safesphere Oct 8 '17 at 16:37
  • $\begingroup$ Sorry to bother you again. But could you tell me what method you used to get integration of the left side (where it results in arctan)? Thank you. $\endgroup$ – Dan Skyler Oct 9 '17 at 11:38
  • $\begingroup$ Not a bother, you can ask anything any time. I don't have time right now to provide the steps, but should be able to do so in a day or two. In general though, finding an integral is not guaranteed, even if it exists. So it is sometimes a bit of art, intuition, and luck in addition to using predefined methods and procedures. The only strict rule is that the integral is correct, if differentiating it gives you the same function that you integrated. So to verify if this result is correct, you can differentiate it and after simplifications you should get the function that was under the integral. $\endgroup$ – safesphere Oct 9 '17 at 18:40
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    $\begingroup$ @DanSkyler Posted the integration steps. Please do not hesitate if you have any questions. $\endgroup$ – safesphere Oct 11 '17 at 5:08
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Let us start with $$\ddot r = -\frac{GM}{r^2}$$

Now multiply with $\dot r$: $$\dot r \ddot r = -\frac{GM \dot r}{r^2}$$

The antiderivatives are $$\frac12 \dot r^2 = \frac{GM}{r} + C$$ where $C$ is some constant.

Thus, $$\dot r = \pm \sqrt{\frac{2GM}{r} + 2C}$$ where sign $+$ is valid for a rising body and sign $-$ is valid for a falling body.

Setting $r = R+h$ then gives $$\dot h = \pm \sqrt{\frac{2GM}{R+h} + 2C}$$

If $\dot h_0=0$ we have the condition $$0 = \sqrt{\frac{2GM}{R+h_0} + 2C}$$ i.e. $$2C = -\frac{2GM}{R+h_0}$$ so we end up with $$ \dot h = \pm \sqrt{\frac{2GM}{R+h} - \frac{2GM}{R+h_0}} = \pm \sqrt{2GM} \sqrt{\frac{1}{R+h} - \frac{1}{R+h_0}} \\ = \pm \sqrt{2GM} \sqrt{\frac{h_0-h}{(R+h)(R+h_0)}} \approx \pm \sqrt{\frac{2GM}{R^2}} \sqrt{h_0-h} \\ = \pm \sqrt{2g(h_0-h)} $$

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