Associated Topics || Dr. Math Home || Search Dr. Math

### Gravity

```
Date: Tue, 1 Nov 1994 20:03:21 -0500 (EST)
From: Boats
Subject: Gravity

Hello Dr. Math,
I teach a course titled The Non-Western World.  Across the hall from
me there is a class of Math Superstars and I'm going to pass on your
service to them.  Meanwhile, I have a question.  When space ships get
close to earth, gravity seems to effect the occupants really fast.  Why
isn't the effect of gravity a gradual occurence?  Does it have anything
to do with that law about any two bodies in the universe affect each
other... and does it have to do with inverse relationships?

Thank you

Fred Hunt   Internet: hunthunt@nando.net   Raleigh, North Carolina
West Millbrook Middle School
```

```
Date: Tue, 1 Nov 1994 20:28:10 -0500
From: The Swat Team
Subject: Re: Gravity

Hi! I'm Heather, and I'm a member of the swat team.  Thanks for writing.

I'm not entirely positive, because it's been a while since I've taken
physics, which is what your question refers to, but I'm thinking that the
answer to your question is that the force (F) between two objects is equal
to the mass of the first (M1) times the mass of the second (M2) divided by
the distance between the two(r) squared, all times some defined constant
that I can't remember (G)...

(M1*M2)
F = (G) --------
r^2

Gravity would affect the occupants quickly because of the r^2 in the
bottom.  As the distance between the two objects decreases, the force
increases exponentially.  I'm sending this to the other Swat team members,
so if there's a problem with this explanation, someone will help.

Date: Wed, 2 Nov 1994 09:20:50 -0500
Subject: Re: Gravity

The inverse square law has little to do with this effect.
Let r be the distance of the rocket from the center of
the earth.  The gravitational force on the rocket is
GMm/r^2, as you say.  At the surface of the earth we have
r^2 = (6.37x10^3 km)^2 = 4.06x10^7 km^2.
In an orbit 200 km above the surface of the earth, which
is well above the point where atmospheric heating is
important, we have r^2 = (6.57x10^3 km)^2 = 4.32x10^7 km^2.
The difference between the two is only about 6%.

The real reason (I think) has to do with the exponential
increase in density of the atmosphere as the rocket
approaches the earth.  The force the astronauts feel
comes from the rapid deceleration of the rocket as it descends
due to the exponentially increasing frictional force of the
atmosphere.

A very important point:  When you are inside a rocket,
as long as it is in free fall you cannot tell that you
are in a gravitational field -- you feel weightless.
(Note that a rocket in orbit beyond the atmosphere
is in free fall.)

The only time you feel heavy is when the rocket
is accelerated in a way the gravity would not produce;
i.e., when the rocket fires its engines, when the
rocket encounters a frictional force, or when the
rocket is sitting on the earth and thereby being prevented
from falling through it.

Nice question!

Josh
```
Associated Topics:
High School Physics/Chemistry

Search the Dr. Math Library:

 Find items containing (put spaces between keywords):   Click only once for faster results: [ Choose "whole words" when searching for a word like age.] all keywords, in any order at least one, that exact phrase parts of words whole words

Submit your own question to Dr. Math
Math Forum Home || Math Library || Quick Reference || Math Forum Search