# Homework Help: 2 questions regarding circular motion and the law of gravity

1. Jan 2, 2007

### Koolaid

hello i am new to the forums. But I have been working on a packet of problems and I got them all done but i have been stuck on these two for a good hour. I have tried to search to internet for possible equations i could use but can't find any that will work or maybe i am just over looking them so i decided to post here and see what some other peoples thoughts are. But I am looking for some guidance on what equations to use.

1. The problem statement, all variables and given/known data

An Earth satellite is orbiting at a distance from the Earth's surface equal to one Earth radius (4000miles). At this location, the acceleration due to gravity is what factor times the value of g at the Earth's surface?

-thoughts
i think im ight be using

F = GMm/r@2

where G is the universal gravitational constant.

G = 6.67*10-11 N-m2/kg2

and F = mg

2. The problem statement, all variables and given/known data

What angular velocity (in revolutions/second) is needed for a centrifuge to produce an acceleration of 1000g at a radius arm of 15cm?

-thoughts

none at the moment

Last edited: Jan 2, 2007
2. Jan 2, 2007

### neutrino

Hi, koolaid. Welcome to PF!

Here are a couple of hints:

#1: Usually, how would you go about finding g at a point at some distance from an object?

#2 What kind of acceleration does the object undergo?

3. Jan 2, 2007

### Koolaid

well for #1 i updated my post

for #2 wouldn't be tangetal acceleration so [tangetal acceleration = radius * angular acceleration] (a=r*a) right?
or would it be centripetal acceleration i am not sure [ centripetal acceleration = tangental velocity squared / radius] (a=v^2/r)

Last edited: Jan 2, 2007
4. Jan 2, 2007

### neutrino

Well, what you've provided is the force on an object of mass 'm' (or 'M') due to the other. The acceleration would be GM/r2 (assuming M is the Earth's mass). You have been provided the distance (from the centre) as twice the Earth's radius. Use that to find that to find the acceleration as a multiple of g (at the surface of the Earth).

Since it has not been mentioned that the speed is changing, it mustrefer to the centripetal accelaration. Write it down in terms of the angular velocity.

5. Jan 2, 2007

### Koolaid

#1. still don't know g or m cause you only have G = 6.67E-11 and r = 4000mi = 6.437E6 m then multiply by 2 = 1.2874E7 m .

#2. a=rw [ centripetal acceleration = radius * angular velocity]

Last edited: Jan 2, 2007
6. Jan 2, 2007

### neutrino

You need not know the values. Moreover, m is not needed since the answer requires you to find only the acceleration due to gravity.

g(r) = GM/r^2

r = 2R , where R is Earth's radius. Therefore,

g(r) = GM/(2R)^2
...
Can you do it from here?
...

The magnitude of centripetal acceleration = w^2*r

7. Jan 2, 2007

### Koolaid

i am still lost.. i am just not really making the "connection" for both

8. Jan 2, 2007

### HallsofIvy

You "need" practically none of the values you have given because the question asks "what factor times the value of g at the Earth's surface". That is, the acceleration at twice the distance from the center of the earth is what number times the acceleration at the surface of the earth? You are really only asked how gravitational accelaration depends on distance from the center of the earth. You don't need to know G, you don't need to know the mass of the earth, you don't need to know the radius of the earth. And, although you must use it, you don't even need to know g! (g, the acceleration due to gravity at the earth's surface is 9.81 m/s2.)

(And be careful not to confuse "G" with "g"!)

You should know the formula: $a= R \omega^2$ where a is the accelleration, R is the radius arm and $\omega$ is the angular velocity (in radians/second- there are $2\pi$ radians in one revolution).

Last edited by a moderator: Jan 2, 2007
9. Jan 2, 2007

### Koolaid

ok i got the first question all done.

but for question 2. i am not sure how to incorporate the 1000 g into the problem.

the radius also needs to be in meters os its .15 meters

10. Jan 2, 2007

### Hootenanny

Staff Emeritus
1000g means one thousand times the acceleration due to gravity [on earth], i.e. $1000 \times 9.81$. So, to rephrase the question, what is the angular velocity required such that the centripetal acceleration is 9810m/s2?

11. Jan 2, 2007

### Koolaid

wow those were simple now that I think about it lol… well now i understand them.. thank you to everyone that helped