• Fr33Fa11
In summary, the conversation revolved around an attempt to calculate the time it takes for a small object to free fall towards a large object, taking into account the increasing gravitational acceleration as it gets closer. The initial approach was to set up a differential equation, but after difficulties, the poster tried using change in kinetic energy. There was discussion about the validity of using a particular form of the differential equation and a suggestion to also consider potential energy. The conversation ended with the poster mentioning their success in solving the problem and asking for a link to previous discussions on the topic.
Fr33Fa11
I have been trying to calculate the time it takes for an object of negligible mass to free fall towards an object with a large mass, taking into account that the gravitational acceleration experienced by the small object increases as it moves closer to the big object.
The first thing I tried was setting up a diffeq:

n''=GM/(h-n)^2
Where G is the gravitational constant, M is the mass of the object, h is the initial height, n is the distance traveled, and ' denotes a derivative. After trying for a few days to solve this equation for n(as a function of t, time), I gave up and tried to solve it using change in kinetic energy, which also didn't work. Any ideas? Thanks.

This problem has been discussed somewhere on this forum, but I don't remember the title of the thread. Anyway, the differential equation (not exactly yours, but basically the same idea) has been solved. I'll give you a hint:

$$\frac{d^2x}{dt^2}=v_x\frac{dv_x}{dx}$$

You should prove this to yourself. (because it is actually not always true, so you should understand why it is true for your particular problem)

I'm not sure I understand why that is true. Is v(x) velocity with respect to x, where x is distance traveled? Is v(x) just an arbitrary name for a function which fits the relationship?

Yes, v_x (not v(x), BTW) is the velocity component in the x direction:

$$v_x\equiv{}\frac{dx}{dt}$$

You can replace x with whatever coordinate you want, but wierder coordinates will require more care in determining when this relationship is true. Are you familiar with the chain rule?

sure, f(g(x))'=f'(g(x))g'(x)
Ok, so far I am following you. The expression rewrites the second derivative of position(acceleration) as the first derivative of position(velocity) times the derivative of velocity with respect to position. Using gravitational potential energy I have figured out the equation for velocity with respect to space, and I know how to take the derivative of that. That leaves just a single order diffeq. A single order differential equation of the form a'=a*k should have a general solution of the form e^(tk), so then acceleration should equal ce^kt where k is the derivative of velocity with respect to position right?

Fr33Fa11 said:
A single order differential equation of the form a'=a*k should have a general solution of the form e^(tk), so then acceleration should equal ce^kt where k is the derivative of velocity with respect to position right?
Are you sure it has that form?

Hi, Fr33Fa11!
Your attempt to use change in kinetic energy should take into accout the change in potential energy, equal to GMm/(h - n) - GMm/h. Basing on it you can find the velocity dependence on distance: V = f(n). The time of flight can be found by integration of the expession dn/f(n).

Last edited:
I've solved it, and I have calculated the constants that come out of the problem. Do you know if this has been done(it would be really cool if it hasn't)

Fr33Fa11 said:
Do you know if this has been done.
I thought we already said that. Sorry, no Nobel prize today.

I was expecting that something like this would have been done before. Do you have a link to the paper/post where it was solved, I want to see if they have a faster way of solving it, because my work is kind of messy, as is the answer.

They did not arrive at a fully solved solution. I'm also trying to solve it for any number of dimensions.(Where GM/r^(n-1)) is the formula for gravitational acceleration with any number of dimensions.

1. What is freefall in a varying gravitational field?

Freefall in a varying gravitational field refers to the motion of an object as it falls towards the surface of a planet or other object, where the strength of the gravitational force changes with distance from the object's center. This results in a non-uniform acceleration and a curved path of motion.

2. How does the strength of gravity affect the time for freefall?

The strength of gravity directly impacts the time for freefall in a varying gravitational field. The stronger the gravitational force, the faster an object will accelerate towards the ground and the shorter the time for freefall will be. Conversely, a weaker gravitational force will result in a slower acceleration and a longer time for freefall.

3. What factors can influence the time for freefall in a varying gravitational field?

The time for freefall can be influenced by various factors, such as the mass of the falling object, the distance from the center of the object it is falling towards, and the shape and density of the object. In addition, external forces, such as air resistance, can also affect the time for freefall.

4. How can the time for freefall in a varying gravitational field be calculated?

The time for freefall in a varying gravitational field can be calculated using the equation t = √(2h/g), where t is the time, h is the initial height of the object, and g is the acceleration due to gravity. This equation assumes a uniform gravitational field, so it may not be accurate for objects falling towards an object with a varying gravitational field.

No, the time for freefall will vary on different planets due to differences in their gravitational fields. For example, the time for freefall on Earth is approximately 9.8 seconds, while on the moon it is only 1.6 seconds. This is because the moon has a weaker gravitational pull compared to Earth.

• Mechanics
Replies
12
Views
1K
• Mechanics
Replies
10
Views
2K
• Mechanics
Replies
8
Views
713
• Mechanics
Replies
27
Views
11K
• Mechanics
Replies
51
Views
1K
• Special and General Relativity
Replies
1
Views
506
• Mechanics
Replies
21
Views
1K
• Special and General Relativity
Replies
12
Views
2K
• Introductory Physics Homework Help
Replies
4
Views
1K
• Classical Physics
Replies
16
Views
1K