# Freefall from a great height, changing g

• bitrex
In summary, the equation for calculating the gravitational force between two objects would change depending on whether the object was in freefall or not. The equation would be different because the force would be based on the mass and acceleration of the object, rather than just its height.
bitrex
This isn't a homework problem exactly - I'm attempting to teach myself introductory physics and there's a question that's bugging me. The kinematics equations that I've seen so far for falling bodies all assume that little g is a constant, i.e. the distance of the fall of the object from its initial height above the Earth is small enough that delta g is zero. My question is how the equation would change for freefall where the starting height was far enough that little g was changing during the fall?

## Homework Equations

y(t) = 1/2gt^2, g = G*M1*M2/r^2

## The Attempt at a Solution

My best guess is that instead of using little g as a constant, i'd integrate F = G*M1*M2/r^2 over the distance traveled and somehow plug that into the equation instead, but my knowledge of both calculus and physics is too shaky to know if I'm on the right track. Can anyone help?

Yes, that's (almost) exactly right- the gravitational force is given by F= -GM1M2/r2 (don't forget the "-". We are measuring r upward but the force is downward). Since we still have "force= mass*acceleration", if M2[/sup] is the mass of the falling object, then acceleration = -GM1/r2. Acceleration is, of course, the derivative of speed with respect to time so we have to integrate, not with respect to the distance but with respect to time. Since the distance does appear in the formula, we have the "differential equation"
$$\frac{dv}{dt}= -\frac{GM_1}{r^2}$$

One thing that makes that complicated is that it is the derivative of v on the left side but we have only r on the right. Since v= dr/dt, we have to use the chain rule to write dv/dt= (dr/dt)(dv/dr)= vdv/dr. Now the equation is
$$v\frac{dv}{dr}= -\frac{GM_1}{r^2}$$

That is what is called a "separable equation" since we can "separate" the variables v and r:
$$v dv= -\frac{GM_1}{r^2}dr$$
and then integrate both sides:
$$\frac{1}{2}v^2= \frac{GM_1}{r}+ C$$
Giving the initial height and speed we could solve for the constant C.

Once we've done that, since v= dr/dt, we have
$$dr/dt= \sqrt{2\frac{GM_1}{r}+ C}$$

While that is also a "separable" equation and can be written as
[tex]\frac{dr}{\sqrt{2\frac{GM_1}{r}+ C}}= dt[/itex]
the left side does not have an "elementary" integral. It gives what is called an "elliptic integral" (obviously related to the elliptic orbits of planets) though special cases can be easy to solve.

Ah, that helps a lot. Thank you!

## What is freefall?

Freefall refers to the motion of an object falling under the sole influence of gravity. It is a type of motion where the only force acting on the object is its weight.

## What is a great height?

A great height is a subjective term and can vary depending on the context. In the study of freefall, it usually refers to heights that are significantly higher than the height of an average person, such as jumping off a tall building or cliff.

## How does changing the acceleration due to gravity affect freefall?

The acceleration due to gravity, denoted as "g", is a constant value on Earth and affects the rate at which objects fall. Increasing or decreasing g will change the speed and duration of freefall for an object. For example, a higher g will result in a faster and shorter freefall, while a lower g will result in a slower and longer freefall.

## What factors can affect freefall from a great height?

The main factor that affects freefall from a great height is the acceleration due to gravity. Other factors that can influence freefall include air resistance and the shape and weight of the object.

## Is freefall from a great height dangerous?

Yes, freefall from a great height can be dangerous and potentially fatal. The speed at which an object falls can cause significant impact upon landing, and the height of the fall can also increase the risk of injury. It is important to take proper safety precautions when conducting experiments or activities involving freefall from a great height.

• Introductory Physics Homework Help
Replies
3
Views
1K
• Introductory Physics Homework Help
Replies
4
Views
1K
• Introductory Physics Homework Help
Replies
7
Views
2K
• Introductory Physics Homework Help
Replies
18
Views
1K
• Introductory Physics Homework Help
Replies
7
Views
1K
• Introductory Physics Homework Help
Replies
30
Views
1K
• Introductory Physics Homework Help
Replies
6
Views
2K
• Introductory Physics Homework Help
Replies
5
Views
1K
• Introductory Physics Homework Help
Replies
13
Views
2K
• Introductory Physics Homework Help
Replies
1
Views
1K