# Gravitation qns

1. Aug 9, 2008

### tianyi.tan

Hello.. some stuff from my tutorial i'm confused about, pls help! thanks:)
1) Why does the moon not fall and hit the earth? Since both exert attractive forces on each other, there should be some kind of acceleration towards a common point.. ><
2)When the earth and moon are held at rest at their present separation and released to move, would the moon orbit the earth? Or would it then fall to the earth?
3) "Despite its mass and size, a typical neutron star can rotate about its own axis at a high speed. What is the fastest rate of rotation of a neutron star such that matter still remains at its surface? Assume that the radius of the neutron star is 20.0km & its mass is 3.98 x 10^30kg." What concepts should I use to solve this problem? thx :D:D

2. Aug 9, 2008

### Staff: Mentor

1. They are constantly falling (being accelerated) toward each other. But they are also moving sideways wrt each other - fast enough that they never hit each other. That's what an orbit is.

2. If they had no tangential velocity, they would fall straight together like a ball dropped straight down from your hand. But they have a tangential velocity, like a ball thrown sideways with respect to the earth.

3. Are these homework? Did you learn about centripedal acceleration yet? There is an equation for it...

3. Aug 9, 2008

### tianyi.tan

hi, thanks so much!
regarding (3), yupp I have learnt about centripetal acceleration, but I'm not sure what they mean by the matter staying on the surface. Am I supposed to equate v2/r to the gravitational field strength? :S

4. Aug 10, 2008

### quark1005

1. The force of attraction is perpendicular to the velocity of the moon's orbit - ie the moon is travelling tangential to the orbit, assuming a circular orbit. Thus there is no component of the force in the direction of motion. The force remains perpendicular to the motion as it moves so it simply changes the direction of the moon's motion. This is an example of uniform circular motion. The acceleration is always directed towards the centre of the Earth (approximately) no matter where the moon is.

2. In this case the moon would move straight towards the earth because it does not have an initial motion perpendicular to the gravitational force.

3. We know
$$a=\frac{GM}{r}$$ (Newton's Law of Gravitation)

and $$a = \frac{v^{2}}{r}$$ (centripetal acceleration)

thus

$$\frac{Gm}{r^{2}} = \frac{v^{2}}{r}$$

$$v= \sqrt{\frac{Gm}{r}}$$ gives the tangential speed of the particles

$$= \sqrt{\frac{6.67e-11*3.98e30}{20e3}}$$

$$= 1.152 *10^8 m/s$$

This is the tangential speed of the outer particles, assuming the star's mass is concentrated at its centre.

Circumfrence $$= 2pir = 1.257 * 10^5m$$

Thus $$1.152 *10^8/1.257 * 10^5 = 917$$ revolutions per second required.

Last edited: Aug 10, 2008