# Free fall equations with realistic, altitude-dependent g?

• xpell
In summary: I don't know, post #20 or so, my head was already spinning).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... I don't know, post #20 or so, my head was already spinning).
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.

xpell said:
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.

Chestermiller said:
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)?

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

Chestermiller said:
scottdave said:
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.

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
scottdave said:
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.

xpell said:
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.

Chestermiller said:
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?

xpell said:
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
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##

Chestermiller said:
##r_0## is defined as the starting radius
##r## is defined as the current 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 going to 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!

xpell said:
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 going to 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

xpell
scottdave said:
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):

Last edited:
scottdave

## 1. What is the formula for calculating free fall with altitude-dependent gravity?

The formula for calculating free fall with altitude-dependent gravity is: F = mg(h/R)^2, where F is the force of gravity, m is the mass of the object, g is the standard acceleration due to gravity, h is the altitude, and R is the radius of the Earth.

## 2. How does altitude affect the acceleration due to gravity?

As altitude increases, the acceleration due to gravity decreases. This is because the distance between an object and the center of the Earth increases, resulting in a weaker gravitational force. The formula for calculating this decrease is g(h) = g(0) * (R/(R+h))^2, where g(0) is the standard acceleration due to gravity at sea level, R is the radius of the Earth, and h is the altitude.

## 3. How do you calculate the time it takes for an object to reach the ground during free fall with altitude-dependent gravity?

The equation for calculating the time it takes for an object to reach the ground during free fall with altitude-dependent gravity is: t = √(2h/g(h)), where t is the time, h is the altitude, and g(h) is the acceleration due to gravity at that altitude.

## 4. Can altitude-dependent gravity affect the trajectory of a falling object?

Yes, altitude-dependent gravity can affect the trajectory of a falling object. As the acceleration due to gravity decreases with altitude, the object will experience a slower rate of acceleration and therefore a longer flight time. This can result in a longer horizontal distance traveled.

## 5. How does air resistance affect free fall equations with altitude-dependent gravity?

Air resistance can affect free fall equations with altitude-dependent gravity by decreasing the acceleration due to gravity at higher altitudes. This is because air resistance acts in the opposite direction of gravity, slowing down the object's fall. To account for air resistance, a drag coefficient can be added to the equation, resulting in a more accurate calculation of the object's fall time and distance.

Replies
1
Views
1K
Replies
3
Views
998
Replies
11
Views
2K
Replies
8
Views
2K
Replies
8
Views
5K
Replies
7
Views
9K
Replies
2
Views
2K
Replies
4
Views
2K
Replies
8
Views
5K
Replies
1
Views
1K