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Let's say we are throwing an object from the surface of the earth, this object reaches 70,000km with initial velocity of $10713 \mathrm{m}/\mathrm{s}$ until it reaches the peak high , the g value at 70,000km is $0.068 \mathrm{m}/\mathrm{s}^2$.

Now my question is: How to calculate the falling time of the object from the peak height (70,000km) until it reaches the ground? Acceleration is not constant here. sketch

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    $\begingroup$ Hello! I have edited your question using MathJax (LaTeX) math typesetting. For future questions, you can refer to MathJax basic tutorial and quick reference. Thanks! $\endgroup$ – Jonas Apr 20 at 9:17
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    $\begingroup$ Assuming no atmosphere? $\endgroup$ – JAlex Apr 20 at 13:34
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    $\begingroup$ I would try to use Kepler laws first. What you describe is an orbit. $\endgroup$ – fraxinus Apr 20 at 18:01
  • $\begingroup$ Is 70K km from surface of earth, the highest altitude, hard to assign clear points with velocity and attitude. $\endgroup$ – marshal craft Apr 20 at 18:12
  • $\begingroup$ Okay and it thrown STRAIT up. $\endgroup$ – marshal craft Apr 20 at 18:13
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Even when acceleration is not constant one can apply conservation of energy.
Let the gravitation potential energy of the earth-object system at the highest point be $-C_1$.

By conversation of energy

$$\frac{-GMm}{x} + \frac{mv^2}{2} = -C_1\tag{1}$$

(where $G$ is universal gravitational constant; $M$ is mass of earth; $m$ is mass of object; $v$ is speed of object; $x$ is distance from center of earth)

From the above equation one can calculate relation between speed and distance from earth

and $$\frac{-\mathrm{d}x}{\mathrm{d}t} = v\tag{2}$$

("$-$" sign is used because as $x$ is decreasing $\frac{-\mathrm{d}x}{\mathrm{d}t}$ is negative)

Using equations $(1)$ and $(2)$ one can find time taken to reach the surface of the earth

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  • $\begingroup$ Edits are encouraged if someone can make my solution more clear by typesetting etc. $\endgroup$ – Dheeraj Kumar Apr 20 at 10:09
  • $\begingroup$ okay i understand the first part where u find the C1 , but im not quite sure what to do after ? i didn't understand the part of -dx/dt thing $\endgroup$ – Majd Apr 20 at 10:14
  • $\begingroup$ if you know a bit about differential calculus, dx/dt is the derivative of x with respect to time and it denotes the instantaneous change in displacement wrt time which is the instantaneous velocity. $\endgroup$ – Dheeraj Kumar Apr 20 at 10:19
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    $\begingroup$ @Majd if you don’t know calculus then you cannot do this. Calculus was invented specifically for this type of problem $\endgroup$ – Dale Apr 20 at 11:15
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    $\begingroup$ @DheerajKumar I assume you mean $$-\frac{GMm}{x},$$ which is the potential energy, otherwise your first equation is not dimensionally correct... $\endgroup$ – Philip Apr 20 at 21:16
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Here I will explain only an alternative procedure to compute the fall time from the maximum height. However, I have to say that the other answer that uses energy is much simpler.

You are given the acceleration as a function of the position and not of time. Let me use $r$ for the radial coordinate from the center of the Earth. You can find the velocity as a function of the position by integration \begin{equation} \int_{r_0}^{r} a(r')\,\mathrm{d}r' = \int_{r_0}^{r} \frac{\mathrm{d}v(r')}{\mathrm{d}t}\,\mathrm{d}r' = \int_{v_0=v(r_0)}^{v(r)} v'\,\mathrm{d}v'= \frac{1}{2}v^2(r) - \frac{1}{2}v^2_0, \end{equation} where I defined $v'=v(r')=\mathrm{d}r'/\mathrm{d}t$.

Therefore the velocity as a function of the radial coordinate is \begin{equation} v(r) = \pm\sqrt{2\int_{r_0}^{r} a(r')\,\mathrm{d}r' + v_0^2}. \end{equation} OBS. in your case $v_0 = 0$ and $a(r) = GM / r^2$.

If you remember that the velocity is the derivative of the position in time, you have that \begin{equation} \frac{\mathrm{d}r}{\mathrm{d}t} = \pm\sqrt{2\int_{r_0}^{r} a(r')\,\mathrm{d}r' + v_0^2} \quad \rightarrow \quad \mathrm{d}t = \pm\frac{\mathrm{d}r}{\sqrt{2\int_{r_0}^{r} a(r')\,\mathrm{d}r' + v_0^2}} \end{equation}

You can integrate this last expression (neglecting the case where the sign is negative) from your initial and final times $t_{fall}$ to get
\begin{equation} t_{fall} = \int_0^{t_{fall}}\mathrm{d}t = \int_{r(t=0)}^R\frac{\mathrm{d}r}{\sqrt{2\int_{r_0}^{r} a(r')\,\mathrm{d}r' + v_0^2}} \end{equation} which is the result you are looking for. Here I used $R$ for the Earth radius.

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    $\begingroup$ Note that you are using energy in this solution, just implicitly. If you multiply both sides of your original equation by $m$, the left-hand side is the work done by the gravitational force (i.e., $-\Delta U$) while the right-hand side is the change in the kinetic energy. $\endgroup$ – Michael Seifert Apr 20 at 13:50
  • $\begingroup$ @MichaelSeifert you are absolutely right $\endgroup$ – Davide Dal Bosco Apr 20 at 13:53
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If you want to figure this out "without calculus", there's a way you can get a good numerical approximation using a computer.

If you take a "sufficiently small" interval of time $\tau$, then the acceleration and the velocity of the object will be approximately constant over that time period. This means that you can use the uniform acceleration equations over that time period. For example, let's suppose that you know what the initial position $x(0)$ and the initial velocity $v(0)$ are. Assuming that you know the relationship between $x$ and $F$, this means that you can infer the acceleration at time $0$ as well. Applying the uniform acceleration equations, we have $$ v(\tau) \approx v(0) + a(0) \tau \qquad x(\tau) \approx x(0) + v(0) \tau + \frac{1}{2} a(0) \tau^2 \approx x(0) + v(0) \tau $$ In the second equation, we can drop the second term because we are assuming that $\tau$ is a "small" amount of time, and a term that is a small number squared is an even smaller number.

These equations get us to $x(\tau)$ and $v(\tau)$. But we can then "reset" the problem and use it as our new starting time, to approximate the position and velocity at $t = 2 \tau$: $$ v(2\tau) \approx v(\tau) + a(\tau) \tau \qquad x(2\tau) \approx x(\tau) + v(\tau) \tau $$ We can repeat this process over & over, iteratively, to find out $x$ and $v$ at $3\tau$, $4\tau$, $5\tau$, and so on, until we have a picture of how the object behaves over a long period of time. In particular, you could keep repeating this process until your object hits the ground, i.e., until $x = R$ for some value of $t$.

Of course, you wouldn't want to do all of this gruntwork by hand. (Though people, often women, did do such gruntwork by hand in the past. A "computer" used to be an occupation, not a device.) But if you're familiar with a programming language, or even if you have a passing familiarity with Excel, you can probably get a computer to do these calculations for you in pretty short order.

What I have described above is what's called the Euler method for numerically solving differential equations. It is perhaps the easiest method to understand. It is also perhaps the worst possible method of its type for solving such equations, since the errors in the approximations are relatively large and in some circumstances can even grow with time. If $\tau$ is small enough, this is generally not a problem, but that means that you have to take a lot of really small time steps to get a solution that spans an interesting amount of time. There are much more sophisticated algorithms out there, and entire libraries of computer functions devoted to solving them. If you understand the Euler method and are interested in using more sophisticated & accurate techniques to solve similar problems, there are plenty out there for you to research.

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Change in kinetic energy is due to work done

$$ \tfrac{1}{2} v(x)^2 - \tfrac{1}{2} v_0^2 = \int_{x_0}^{x} a(x)\,{\rm d}x $$

and since acceleration is a function of position $a(x)$, the above integral wil give you the velocity-distance relationship $v(x)$.

Now time to reach distance is a function of the velocity curve

$$ t = \int_{x_0}^{x} \frac{1}{v(x)} \,{\rm d} x $$


On the other hand if you had acceleration as a function of speed only $a(v)$ then you use two other integrals

$$ x(v) = x_0 + \int_{v_0}^{v_1} \frac{v}{a(v)}\,{\rm d}v $$

$$ t(v) = \int_{v_0}^{v_1} \frac{1}{a(v)}\,{\rm d}v $$

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  • $\begingroup$ i'm not quite sure how to start , i understood the concept , but im not sure how to use the integrals ... $\endgroup$ – Majd Apr 20 at 15:50
  • $\begingroup$ If $x$ is the height above the surface, then $a(x) =- g\, R^2/(R+x)^2$ $\endgroup$ – JAlex Apr 20 at 18:36
  • $\begingroup$ So basically $$t=\int\limits_{x_i}^{x_f}\frac{1}{\sqrt{v_i^2+2\int\limits_{x_i}^{\lambda}a(x)dx}}d\lambda$$ $\endgroup$ – user5402 Apr 20 at 22:23
  • $\begingroup$ Yes, I think $v_i=0$ which I think simplifies it just a tad. $\endgroup$ – JAlex Apr 21 at 13:25
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I will only outline an answer for now.

The object is on a suborbital trajectory. Classically this just means the objects or it is a 2 body gravitational orbit like earth and moon except the periapsis is lower altitude then the surface of the earth. Describing orbital characteristics is the same.

Now there is usually no time parameterization of the orbit. One usually has to use an approximation method like newton's method, to get the actual angle (true anomaly) from the circle angle (mean anomaly) for times ellapsed past an epoch (known position). This is due a interesting term which involves sin function, and there is no way to algebraically solve.

So basically, object experience known not necessarily constant acceleration is simple. Object orbiting from gravitational field is not so simple. It gets into transcendental functions, non linear pdes etc. That's classical gravity and differential equations for you.

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  • $\begingroup$ Sometimes due to inherent symmetry, you do not need to do this. You CAN describe the x,y, parameterized orbit path pretty easy. It's an ellipse. But finding the times may or may not be able to take advantage of symmetries. Arbitrary points prob not. $\endgroup$ – marshal craft Apr 20 at 18:10
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Integrating twice based on acceleration due to gravity. (Keplar could also be used, but the result is the same). m1 = mass of earth m2 = mass of object, so much smaller than mass of earth it can be ignored

gravitational constant G = 6.6743 x 10^-11

m1+m2 = 5.9722 x 10^24

r is distance between the two objects

v is combined closure rate of the two objects towards a common center of mass.

a is combined acceleration of the two objects towards a common center of mass

initial distance r0 = 76370000

final distance r1 = 6370000

$$ v = dr/dt \\ a = dv/dt \\ a = (dv / dr) (dr / dt) \\ a = v \ dv/dr = -G (m1 + m2) / r^2\\ v \ dv = -G (m1 + m2) dr / r^2 $$ Integrating both sides, with constant v = 0 at r0 $$ 1/2 v^2 = G(m1+m2)/r - G(m1+m2)/r0 \\ v = -\sqrt{2G(m1+m2)/r - G(m1+m2)/r0} \\ v = \frac{dr}{dt} = -\sqrt{\frac{2 \ G\ r0 \ (m1+m2)-2 \ G \ r \ (m1+m2)}{r \ r0}} \\ \frac{- \sqrt{r0 \ r} \ dr} {\sqrt{2 \ G \ r0 (m1 + m2) - 2 \ G \ r \ (m1 + m2)}} = dt \\ {- \sqrt{ \frac{r0}{2 \ G \ (m1 + m2)}}} \ \ \sqrt{\frac{r}{r0 - r}} \ dr = dt \\ $$ Integrate again, using u for substitution. $$ u = \sqrt{\frac{r}{r0-r}} \\ u^2 = \frac{r}{r0-r} \\ r = \frac{r_0 u^2}{1 + u^2} = \frac{r_0 + r_0 u^2 - r_0}{1 + u^2} = \frac{ r_0(1 + u^2) - r_0}{1 + u^2} = r_0 - \frac{r_0}{1 + u^2} \\ dr = \frac{2 \ r0 \ u \ du}{(1 + u^2)^2} \\ when \ r = r0 = 76370000\ m, u0 = \infty \\ when \ r = r1 = 6370000\ m, u1 = \sqrt{0.091} \\ t = \frac{-2 {r0}^{3/2}}{\sqrt{2 G (m1 + m2)}} \ \int_\infty^\sqrt{0.091} \frac{u^{2}}{(1+u^{2})^{2}}du \\ t = \frac{-2 \ {r0}^{3/2}}{\sqrt{2 G (m1 + m2)}} \left [ \frac{1}{2} \left (\tan^{-1}(u)-\frac{u}{1+u^{2}} \right ) \right ] _\infty^\sqrt{0.091} \\ t = \frac{{-r0}^{3/2}}{\sqrt{2 G (m1 + m2)}} \left (.01648041 - \frac{\pi}{2} \right ) \\ t \approx 32240 \ seconds $$

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