# B Free fall equations with realistic, altitude-dependent g? (1 Viewer)

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#### xpell

Hi! I was wondering about an object free-falling to Earth (or wherever) in a vacuum from a very high altitude where g is significantly lower than on the surface: let's say 5,000 km for instance, where g would be 3.08 m/s2 according to the equation gheight = gsurface · (RadiusEarth / (RadiusEarth + Height))2 and will obviously increase during the fall. I have found this for time-dependent non-constant acceleration, but nothing for distance-dependent accelerations. Please, can you help with a couple of motion equations to calculate distance, time and velocity during the fall in this scenario? Thank you in advance!

PS. My knowledge of Math and Physics is close to negligible and I'm not a student, just a 50 year old curious person, but I'm still able to solve an equation if I can find it.

PS2. Please ignore drag and other atmospheric effects for simplicity. I assume that the final velocity in the atmosphere would be the terminal velocity, but not while still in outer space.

#### Chestermiller

Mentor
Hi! I was wondering about an object free-falling to Earth (or wherever) in a vacuum from a very high altitude where g is significantly lower than on the surface: let's say 5,000 km for instance, where g would be 3.08 m/s2 according to the equation gheight = gsurface · (RadiusEarth / (RadiusEarth + Height))2 and will obviously increase during the fall. I have found this for time-dependent non-constant acceleration, but nothing for distance-dependent accelerations. Please, can you help with a couple of motion equations to calculate distance, time and velocity during the fall in this scenario? Thank you in advance!

PS. My knowledge of Math and Physics is close to negligible and I'm not a student, just a 50 year old curious person, but I'm still able to solve an equation if I can find it.

PS2. Please ignore drag and other atmospheric effects for simplicity. I assume that the final velocity in the atmosphere would be the terminal velocity, but not while still in outer space.
Getting the velocity vs distance is pretty easy by applying the principle of conservation of energy. However, getting the distance vs time requires solution of an ordinary differential equation.

This is not the first time that this exact problem has appeared in Physics Forums.

#### xpell

Getting the velocity vs distance is pretty easy by applying the principle of conservation of energy. However, getting the distance vs time requires solution of an ordinary differential equation.

This is not the first time that this exact problem has appeared in Physics Forums.
Hi Chester, thank you very much, I think I would be able to solve such equations. I have searched the forums extensively before posting my question (I also thought that it would be a common question) but I was unable to find it. Would you please be so kind to point me to the appropriate thread(s)?

#### scottdave

Homework Helper
Gold Member
The thread that @Chestermiller pointed to, looks like what you need to solve this type of problem.

#### xpell

The thread that @Chestermiller pointed to, looks like what you need to solve this type of problem.
Thank you both very much, I really appreciate it, but I am reading that thread and I'm already completely lost. As I said, I'm not a specialist or a student, just a worker with a bit of Sunday free time and lots of curiosity. This was a "Sunday thought", I don't have the time or ability to study a full course in Physics to answer it. I was just looking for a set of straightforward equations, as in t = this and v = that, even if it implies solving a couple integrals, just like t = t0 + √(2d/a), then v = v0 + at when the acceleration is constant and the distance is given. I can try and solve them even if they're difficult, but I'm totally unqualified and unable to deduce them myself. Thank you anyway.

#### scottdave

Homework Helper
Gold Member
So we have acceleration = g = G*me/r2. r is the radius to the center of the earth. me is the mass of the earth.
Just know that g is proportional to 1/r2.
acceleration = Δv / Δt → dv/dt in the limit (Calculus). But then v = dr/dt

So we have d2r/dt/r2 is proportional to 1/r2. This can be solved with Calculus to get an expression for r(t). If I have time I'll go back and do the LaTeX for the formulas.

• xpell

#### xpell

So we have acceleration = g = G*me/r2. r is the radius to the center of the earth. me is the mass of the earth.
Just know that g is proportional to 1/r2.
acceleration = Δv / Δt → dv/dt in the limit (Calculus). But then v = dr/dt

So we have d2r/dt/r2 is proportional to 1/r2. This can be solved with Calculus to get an expression for r(t). If I have time I'll go back and do the LaTeX for the formulas.
I'm totally unable to do that, Scott. I'm just able to solve stuff like t or v = integral of this-and-that multiplied by this-other-thing, squared. That's why I needed the straightforward equations. Don't worry, I guess this is beyond what I could study too many years ago, I appreciate your help very much anyway.

#### Chestermiller

Mentor
I'm totally unable to do that, Scott. I'm just able to solve stuff like t or v = integral of this-and-that multiplied by this-other-thing, squared. That's why I needed the straightforward equations. Don't worry, I guess this is beyond what I could study too many years ago, I appreciate your help very much anyway.
An algebraic solution to the problem (of the form you seem to desire) is presented in post #75 of the referenced thread.

#### xpell

An algebraic solution to the problem (of the form you seem to desire) is presented in post #75 of the referenced thread.
Thank you very much, Chester! I guess that's #74? But what do the variables r, r0, u, u0 and u1 mean to properly substitute and calculate, please? (I have followed the entire discussion as hard as I could, but as I said, I got lost halfway and by post #75 I wasn't understanding anything...)

Let's say the object is falling from 5,000 km above the mean Earth radius and it's now at 3,000 km (for instance, choose whatever values you prefer.) How do I substitute that in this solution to calculate its velocity and the elapsed time, please?

#### Chestermiller

Mentor
Thank you very much, Chester! I guess that's #74? But what do the variables r, r0, u, u0 and u1 mean to properly substitute and calculate, please? (I have followed the entire discussion as hard as I could, but as I said, I got lost halfway and by post #75 I wasn't understanding anything...)

Let's say the object is falling from 5,000 km above the mean Earth radius and it's now at 3,000 km (for instance, choose whatever values you prefer.) How do I substitute that in this solution to calculate its velocity and the elapsed time, please?
$r_0$ is defined as the starting radius
$r$ is defined as the current radius
$r_1$ is the final radius
u is defined in post #73 in terms of $r$ and $r_0$
$u_0$ the value of u at $r=r_0$
$u_1$ is the value of u at $r=r_1$

#### xpell

$r_0$ is defined as the starting radius
$r$ is defined as the current radius
$r_1$ is the final radius
u is defined in post #73 in terms of $r$ and $r_0$
$u_0$ the value of u at $r=r_0$
$u_1$ is the value of u at $r=r_1$
Thank you again very much. I honestly thought that this was a common, obvious or even "classical" problem with a "set answer" since there's Calculus because, after all, it's basically the famous Galileo's Tower of Pisa experiment with realistic gravity. I was actually surprised when I didn't find the equations using Google and then I came to ask thinking "sure the experts at PhysicsForums are gonna tell me to learn to use Google better." I didn't even imagine that it would be so complex. I'm really sorry for bothering you and I thank you a lot for indulging me!

#### Chestermiller

Mentor
Thank you again very much. I honestly thought that this was a common, obvious or even "classical" problem with a "set answer" since there's Calculus because, after all, it's basically the famous Galileo's Tower of Pisa experiment with realistic gravity. I was actually surprised when I didn't find the equations using Google and then I came to ask thinking "sure the experts at PhysicsForums are gonna tell me to learn to use Google better." I didn't even imagine that it would be so complex. I'm really sorry for bothering you and I thank you a lot for indulging me!
You have been no such bother. It is our pleasure to help you, and we would be pleased to help you in the future. Helping people understand science is what Physics Forums is all about. Thanks for considering us.

Chet

#### rcgldr

Homework Helper
This can be solved with Calculus to get an expression for r(t).
Nobody here has posted a direct way to do this, at least not yet. What the prior threads have been able to solve is the time for the radius to decrease from r0 to r1, an equation that doesn't appear to be invertible (the equation can't be inverted to calculate radius as a function of time). It would be possible to do something like binary search for r1 for a given r0 and elapsed time, or as suggested by Chestermiller in the prior thread linked to above, create a lookup table (or perhaps use both, use the table to get a starting value for a binary search).

In one of the prior threads, an alternate solution base on Kepler's law is shown, but the final equation is essentially the same (go to post #11 of this thread):

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• scottdave

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