Time of a falling object when the force of gravity isn't constant

[Chestermiller]

I integrated 1/v(r) dr (and v(r) )and I had an expression that I cna't understand what it means.

Let's see what you did. This is exactly what rdgldr did, so we can compare notes, and, if you went wrong, he can point out where.

Chet

 

[Shahar]

a = G * M
v(x) = √(2 *a)/x
1/v(x) = 1/√(2 *a)/x = √x / √2a

∫ 1/v(x) dx =
=∫ √x / √2a dx =
=1/√2a ∫ √x =
=1/√2a * x^1.5/1.5 =
= √2 * x^1.5/3a

I'm not sure it's correct.
And I reread post #27 about this equation but I don't undersatnd how to apply it.
From what I understand this is t, and It doesn't fit when I plug in x1 and x2.

 

[Chestermiller]

a = G * M
v(x) = √(2 *a)/x

This equation is incorrect. It should read:
$$v(x)=-\frac{dx}{dt}=\sqrt{2GM}\sqrt{\frac{1}{x}-\frac{1}{x_0}}=\sqrt{\frac{2GM}{x_0}}\sqrt{\frac{x_0-x}{x}}$$
where x₀ is the starting radius.

Chet

 

[rcgldr]

Go back to post #50 to see how to continue from here.

 

[Shahar]

Go back to post #50 to see how to continue from here.

Ok, I read post #50 a few times and I am lost. I don't understand how the equations contributor to the solution.

 

[rcgldr]

Ok, I read post #50 a few times and I am lost. I don't understand how the equations contributor to the solution.

From post #28, you have this complicated result:

http://www.wolframalpha.com/input/?i=integral+-+sqrt(r0+/+(2+G+M))+sqrt(r+/+(r0-r))+dr

By using the substitution from post #34:
$$- \ \sqrt{\frac{r_0}{2 G M}} \sqrt{\frac{r}{r_0-r}} dr = dt $$
$$ r = r_0 \ sin^2(θ) $$
$$\sqrt{\frac{r}{r_0-r}} = \sqrt{\frac{r_0 \ sin^2(θ)}{r_0-r_0 \ sin^2(θ)}} = \sqrt{\frac{r_0 \ sin^2(θ)}{r_0(1\ -\ sin^2(θ))}} = \sqrt{\frac{r_0 \ sin^2(θ)}{r_0\ cos^2(θ)}} = tan(θ) $$
$$dr = 2\ r_0 \ sin(θ)\ cos(θ) dθ$$
$$\sqrt{\frac{r}{r_0-r}} dr = (tan(θ)) \ (2 \ r_0 \ sin(θ)\ cos(θ) dθ) = 2 \ r_0 \ sin^2(θ) dθ $$
You end up with this equation to integrate:
$$-\sqrt{\frac{r_0}{2 \ G \ M}} \ (2 \ r_0 \ sin^2(θ)) dθ = dt $$
and you need to convert r = r₀ or r₁ to θ = θ₀ or θ₁
$$θ = {sin^{-1}\left ( \sqrt{\frac{r}{r_0}} \right )} $$

 

[Chestermiller]

rcgldr:
It looks like Shahar is not yet advanced to the point where he can complete this integration. Why don't you complete the analytic solution and then calculate the time to fall to the planet? It would be interesting to compare this with the value of 2000 sec that Shahar estimated.

Shahar: The solution to this problem can be obtained graphically, by finding the area under the curve of 1/v(r) vs r. Could you please plot a graph of 1/v(r) (units of sec/meter) as a function of r (meters) and show us what you get? If you could, that would be great.

Chet

 

[rcgldr]

$$dt = -\sqrt{\frac{r_0}{2 \ G \ M}} \ (2 \ r_0 \ sin^2(θ)) dθ$$
$$t = -\frac{{r_0}^{3/2}}{\sqrt{2 \ G \ M}} \int_{θ_0}^{θ_1}2 \ sin^2(θ) dθ = -\frac{{r_0}^{3/2}}{\sqrt{2 \ G \ M}} \left [ θ \ -\ sin(θ)cos(θ) \right ]_{θ_0}^{θ_1}$$

 

[Shahar]

$$dt = -\sqrt{\frac{r_0}{2 \ G \ M}} \ (2 \ r_0 \ sin^2(θ)) dθ$$
$$t = \int_{θ_0}^{θ_1} -\sqrt{\frac{r_0}{2 \ G \ M}} \ (2 \ r_0 \ sin^2(θ)) dθ = -\sqrt{\frac{r_0}{2 \ G \ M}} \left [ 2\ r_0\ (\frac{1}{2}\ (θ\ -\ sin(θ)cos(θ))) \right ]_{θ_0}^{θ_1}$$
$$t = -\frac{{r_0}^{3/2}}{\sqrt{2 \ G \ M}} \left [ θ\ -\ sin(θ)cos(θ) \right ]_{θ_0}^{θ_1}$$
$$r_0 = {10}^{14}, r_1 = {10}^7 $$
$$θ_0 = \frac{π}{2}, θ_1 = {10}^{-7} $$
$$t = -\frac{{r_0}^{3/2}}{\sqrt{2 \ G \ M}} \left [ θ\ -\ sin(θ)cos(θ) \right ]_{π/2}^{{10}^{-7}}$$

r₀ is 2*10^7. I made a mistake when I misread the question.
And θ₁ is only 0.00031 not 10^-7

 

[rcgldr]

$$r_0 = 2e^7 \ , \ r_1 = 1e^7 $$
$$θ_0 = π/2 \ , \ θ_1 = π/4$$
$$t = -\frac{{r_0}^{3/2}}{\sqrt{2 \ G \ M}} \left [ θ\ -\ sin(θ)cos(θ) \right ]_{π/2}^{π/4}$$
This result now matches the result in post #73:
$$t = -2000 ((π/4 - 1/2) -\ (π/2 - 0)) \simeq 2570.8 \ seconds $$

 

[Shahar]

$$r_0 = 2e^7, r_1 = 1e^7 $$
$$θ_0 = \frac{π}{2}, θ_1 = .000316$$
$$t = -\frac{{r_0}^{3/2}}{\sqrt{2 \ G \ M}} \left [ θ\ -\ sin(θ)cos(θ) \right ]_{π/2}^{.000316}$$
Need to check the math on this part:
$$t \simeq -1999 (.00031 -\ π/2) \simeq 3139 \ seconds $$

I recalculated θ. θ = 0.78
So t = 2528 seconds.

 

[Chestermiller]

I recalculated θ. θ = 0.78
So t = 2528 seconds.

Yes. It's π/4.

Chet

 

[rcgldr]

Using the other substitution:
$$u = \sqrt{\frac{r}{r_0 \ - \ r}} $$
$$r = r_0 - \frac{r_0}{1 + u^2} $$
$$dr = \frac{2 r_0 u \ du}{(1 + u^2)^2}$$
$$t = -\ \frac{ {r_0}^{3/2}}{\sqrt{2 G M}} \int_{u0}^{u1} \frac{2 \ u^{2}}{(1+u^{2})^{2}}du= -\ \frac{ {r_0}^{3/2}}{\sqrt{2 G M}} \left [ \tan^{-1}(u)-\frac{u}{1+u^{2}} \right ] _{u0}^{u1}$$
$$r_0 = 2e7 \ , \ r_1 = 1e7$$
$$u0 = ∞ \ , \ u1 = 1$$
$$tan^{-1}(∞) = π/2 \ , \ ∞ / (1 \ + \ ∞^2) = 0$$
This result now matches the result in post #70:
$$t = -2000 ((π/4 - 1/2) -\ (π/2 - 0)) \simeq 2570.8 \ seconds $$

 

[Shahar]

Using the other substitution:
$$ u = \sqrt{\frac{r}{r_0 \ - \ r}} $$
$$t = -\ \frac{ {r_0}^{3/2}}{\sqrt{2 G M}} \left [ \tan^{-1}(u)-\frac{u}{1+u^{2}} \right ] _{u0}^{u1}$$
r0 = 2e7
r1 = 1e7
u0 = ∞
u1 = 1
$$t \simeq -1999(.2854 - π/2) \simeq 2570$$

This way is much clearer.
I think Ill learn more calculus and then try to get to get to this equation. Or try another simular one( I want to calculate V and t for a free falling object with air resistance
)

 

[rcgldr]

$$ t \simeq 2570.8 $$

I fixed posts #70 and #73.

Or try another simular one( I want to calculate V and t for a free falling object with air resistance)

For this one, the drag related component of acceleration is related to speed^2, not position, so not quite the same. For a vertical drop, the problem can be solved, for the more general case of an object that also has a horizontal component of velocity, you need to use a numerical integration method.

As mentioned before a simpler problem for acceleration based on position is:
acceleration = -x
at time t = 0, x = 0, v = 1 meter / second
You'll be able to solve this and determine x(t), v(t) and a(t).

 

[Chestermiller]

$$ t \simeq 2570 $$

I fixed posts #70 and #73, and now the results are the same, but still not sure since you got 2528.

That's because he used an inaccurate value of 0.78 for π/4. Your answer of 2570 is spot on.

Chet

 

[Shahar]

$$ t \simeq 2570 $$

I fixed posts #70 and #73, and now the results are the same, but still not sure since you got 2528.

$$\frac{{r_0}^{3/2}}{\sqrt{2 G M}} = \frac{{(2e7)}^{3/2}}{\sqrt{2 \times {6.67384e{-11}} \times {1.5e25}}} \simeq 1999$$

For this one, the drag related component of acceleration is related to speed^2, not position, so not quite the same. For a vertical drop, the problem can be solved, for the more general case of an object that also has a horizontal component of velocity, you need to use a numerical integration method.

As mentioned before a simpler problem for acceleration based on position is:
acceleration = -x
at time t = 0, x = 0, v = 1 meter / second
You'll be able to solve this and determine x(t), v(t) and a(t).

I learned a lot from this question and I think,
with a bit more calculus I will be able to find the motion equations with air resistance

 

[rcgldr]

That's because he used an inaccurate value of 0.78 for π/4. Your answer of 2570 is spot on.

I cleaned this up a bit:

Given g = 10 m / s^2 at r = 1e7 m:
g = M G / r^2
G M = r^2 g = (1e7)^2 x 10 = 1e15
2 G M = 2e15
$$\frac{{r_0}^{3/2}}{\sqrt{2 G M}} = \frac{(2e7)^{3/2}}{\sqrt{2e15}} = 2e7 \sqrt{\frac{2e7}{2e15}} = 2e7 \sqrt{\frac{1}{2e8}} = \frac{2e7}{2e4} = 2000$$
$$t = -2000 ((π/4 - 1/2) -\ (π/2 - 0)) = 2000 (π/4 + 1/2) \simeq 2570.7963267949 $$

u = sqrt(r/r0-r) (versus r = r0 sin^2(θ))

This way is much clearer.

I'd recommend going back and looking at the two methods again. Although u = sqrt(r/r0-r) is a typical substitution method, in this case, with a bit of insight, using r = r0 sin^2(θ) replaced the square root component with tan(θ), and integrating 2 sin^2(θ)dθ is simpler than integrating (2 u^2 du) / (1 + u^2)^2 .

 

[PWiz]

So is there no way to express $r$ in terms of $t$ only?

 

[Chestermiller]

So is there no way to express $r$ in terms of $t$ only?

Of course there is. Go to it rcgldr.

Chet

 

[PWiz]

Btw, does this method assume that the planet remains stationary, or is the motion of both bodies included?

 

[PWiz]

I think I've found a neater solution to the problem. If you leave the equation in the form $$t=\int_{r_0}^r \frac{-dr}{\sqrt{\frac{2GM}{r} - \frac{2GM}{r_0}}}$$
and make the substitution $r=\frac{r_0}{sec^2 \theta}$, you get
$$t=r_0 \sqrt{ \frac{r_0}{2GM}} (arccos(\sqrt{\frac{r}{r_0}})+\frac{\sqrt{r} \sqrt{r_0 -r}}{r_0})$$ with no pi or extra -ve signs.
(You can also arrive at this solution by considering the identity $sin(\frac{\pi}{2} - \theta) = cos(\theta)$ . Since the roles of the domain and range are switched with the inverse function, you get the identity $\frac{\pi}{2} - arcsin(\theta) = arccos(\theta)$ and arrive at the solution above.)

And so far, the only progress I've been able to make in using the function $t(r)$ to obtain the function $r(t)$ is the expression the solution in terms of a(n affine?) parameter $\lambda$ :
$$\sqrt{\frac{r}{r_0}}=cos \ \lambda ,$$ and
$$\frac{1}{r_0} \sqrt{\frac{2GM}{r_0}} t = cos \ \lambda (1+ sin \ \lambda)$$
I was thinking about using these Taylor sequences to go further:
$$cos \ \lambda = \sum_{n=0}^∞ \frac{(-1)^n \lambda^{2n}}{(2n)!}$$
$$sin \ \lambda = \sum_{n=0}^∞ \frac{(-1)^n \lambda^{2n+1}}{(2n+1)!}$$
However, this would then mean that the functions $r(t)$, $v(t)$ and $a(t)$ would all have an infinite series, which would look very odd (pardon the parity pun). Is there any other way around this?

On a side note, doesn't the equation $\frac{dr}{dt}=-\frac{dx}{dt}$ assume that $M>>m$ so that $M$ remains stationary relative to $m$?

 

[Shahar]

On a side note, doesn't the equation $\frac{dr}{dt}=-\frac{dx}{dt}$ assume that $M>>m$ so that $M$ remains stationary relative to $m$?

Yes, $M$ remains stationary relative to $m$.
I would be happy to derive an equation for $M$ ~ $m$ so $M$ isn't stationary, but I don't have enough knolige in calculus.

 

[PWiz]

Yes, $M$ remains stationary relative to $m$.
I would be happy to derive an equation for $M$ ~ $m$ so $M$ isn't stationary, but I don't have enough knolige in calculus.

Here, get started :)
http://tutorial.math.lamar.edu/Classes/CalcI/CalcI.aspx

 

[Chestermiller]

I think I've found a neater solution to the problem. If you leave the equation in the form $$t=\int_{r_0}^r \frac{-dr}{\sqrt{\frac{2GM}{r} - \frac{2GM}{r_0}}}$$
and make the substitution $r=\frac{r_0}{sec^2 \theta}$, you get

This is certainly not any "neater" than the substitution I recommended in post #34, $r=r_0sin^2θ$, since, if $r=\frac{r_0}{sec^2 \theta}$, then $r=r_0cos^2θ$, which is essentially the same substitution. Your θ is just π/2 minus my θ. In posts #50, and 66-70, rcgldr employed my substitution, and obtained the same results you have obtained. Your parameter λ is exactly the same as your parameter θ. If you substitute $π/2-θ$ (Chet's θ) for λ in your final equation, you get rcgldr's results.

I was thinking about using these Taylor sequences to go further:
$$cos \ \lambda = \sum_{n=0}^∞ \frac{(-1)^n \lambda^{2n}}{(2n)!}$$
$$sin \ \lambda = \sum_{n=0}^∞ \frac{(-1)^n \lambda^{2n+1}}{(2n+1)!}$$
However, this would then mean that the functions $r(t)$, $v(t)$ and $a(t)$ would all have an infinite series, which would look very odd (pardon the parity pun). Is there any other way around this?

Using a Taylor series expansion is definitely not the way to solve this non-linear algebraic equation for r as a function of t. There are numerous methods available for numerical solution, including the half-interval technique and Newton's method. (Of course, you could say that Newton's method involves expanding to the linear term in a Taylor series, but this is not what I think you had in mind.)

If I wanted r as a function of t, I would just make a table of t(r) vs. r, and plot t as the abscissa and r as the ordinate.

Chet

 

[PWiz]

This is certainly not any "neater" than the substitution I recommended in post #34, $r=r_0sin^2θ$, since, if $r=\frac{r_0}{sec^2 \theta}$, then $r=r_0cos^2θ$, which is essentially the same substitution. Your θ is just π/2 minus my θ. In posts #50, and 66-70, rcgldr employed my substitution, and obtained the same results you have obtained. Your parameter λ is exactly the same as your parameter θ. If you substitute $π/2-θ$ (Chet's θ) for λ in your final equation, you get rcgldr's results.Using a Taylor series expansion is definitely not the way to solve this non-linear algebraic equation for r as a function of t. There are numerous methods available for numerical solution, including the half-interval technique and Newton's method. (Of course, you could say that Newton's method involves expanding to the linear term in a Taylor series, but this is not what I think you had in mind.)

If I wanted r as a function of t, I would just make a table of t(r) vs. r, and plot t as the abscissa and r as the ordinate.

Chet

OK Chet, I guess it was just me trying to avoid pi in the expression. I'd really appreciate if you can give an approximate expression for $r(t)$ though (or is the function implicit?).

 

[Chestermiller]

OK Chet, I guess it was just me trying to avoid pi in the expression. I'd really appreciate if you can give an approximate expression for $r(t)$ though (or is the function implicit?).

The function is implicit. So you really need to solve the non-linear algebraic equation for r(t).

Just make a table with column 1 being r and column 2 being t. Choose values of r in column 1 in equal increments (between the starting value r₀ and the final value at the planet surface). Calculate the corresponding values of t from your equation, and fill them in in column 2. Then plot a graph of r vs t. This is the easiest thing to do. Just be happy that t can be expressed explicitly in terms of r.

Chet