# Really,reall need your help-I'm totally lost

1. Nov 2, 2005

### boris16

hello

The following is really giving me troubles

*What is needed for an object to circle around the earth?
Let me rephrase this...what conditions need to be met for an earth's gravity force to act as centripetal force instead of just pulling an object towards its center?

*When you toss a ball horizontally, the first instant the force is perpendicular to the velocity, so its direction changes. But then it's no longer moving horizontally any more, so gravity is no longer perpendicular to the motion: the ball starts speeding up as it falls.

Thing to point out is that while horizontal component of velocity doesn't change in magnitude, it does change its direction at the brief moment when force of gravity was perpendicular to it.
But after that first moment net force(its vertical component) keeps getting bigger as ball is falling towards the earth

If all earths mass was gathered at earth's center, tossed ball would circle around it.
But why is that? Force is no longer perpendicular to velocity vector, or is it?

I need to know exactly what's going on with regards to force of gravity,its direction on ball if all earth's mass is located at its center

thank you

2. Nov 2, 2005

### Staff: Mentor

>*What is needed for an object to circle around the earth?

The tangential speed of the object needs to be just right. Too fast and it will spiral away from the earth. Too slow, and it will spiral in. There is a correct speed for each radius of a stable orbit. Like how geostationary satelites have an orbit with a period of 24 hours....

The force of the Earth's gravity is pulling the object straight towards the center of the Earth, with a force of G(Mm)/r^2. Recall the formula for the inward-directed acceleration of an object that is in circular motion, and use F=ma to solve for the stable orbital speed as a function of r.

3. Nov 2, 2005

### boris16

You're moving too fast. First I need to get a bigger picture of what is happening. For that I need to know how force of gravity influences velocity vector ( its components ) of a tossed ball ( if all earth's mass located in single point ).

I need to get solid understanding of the basics

4. Nov 3, 2005

### boris16

I'm especially interested in a position of force of gravity if we toss a ball horizontally.
In drawings you always see vertical force perpendicular to horizontal velocity.
But is that really the case?Peerhpas it doesn't make a difference if ball drops down only after 100 m, but what if the ball would drop only after 1000 km?

Wouldn't in that case (where the distance ball travels is that much longer) mean that force of gravity is no longer perpendicular to horizontal velocity,and further the ball goes bigger the angle between horizontal velocity and force of gravity? Need some details on that

5. Nov 3, 2005

### andrevdh

$F_G$ needs to be perpendicular to the velocity of the object during all of it's motion so that all what it does is to change the direction of the velocity of the object, but it's speed stays the same. $F_G$ will therefore cause the object to fall towards the centre of the earth, but while it falls inwards it's velocity carried it such a distance along that it still remains the same distance from the centre of the earth.

6. Nov 3, 2005

### boris16

But F[g] is perpendicular to horizontal velocity vector ( V[h] ) only for a brief instant and then it is no longer perpendicular to it so I don't see how it can change its direction. What's more, since the further away the ball is the bigger the angle between V[h] and F[g], which means that now F[g] has horizontal component F[h] pulling on ball in opposite direction as V[h]

I just don't understand how a ball could go in elliptical orbit

7. Nov 4, 2005

### andrevdh

$$F_G$$ is directed along the line connecting the two objects. The object (m) that orbits the larger one (M) therefore would have moved straight on away from M, but with $$F_G$$ it now also falls inwards towards it. If these two motions are in the right proportion - away and inwards the resulting motion can be such that it stays perpendicular to $$F_G$$ after each interval, since the direction of $$F_G$$ would change slightly after such. I am reffering to centripetal motion with m moving in a circle around M at a constant speed.

8. Nov 5, 2005

### andrevdh

The velocity of a moving object is always tangential to the curve it is moving along. The away displacement is $\Delta s$ and the inwards displacement is the $\Delta r$ bit.

Last edited: Nov 29, 2006
9. Nov 6, 2005

### boris16

I must say I'm still sort of confused about this stuff :(

You're talking about centripetal forces here right?

What do $\Delta s$ and $\Delta r$ represent? Distance over period of time perhaps or change of velocity vector ...?

I'm most interested in elliptical orbit tossed ball would travel in if all earth's mass was located at earth's center- how forces make the object travel in that path.

10. Nov 7, 2005

### andrevdh

I am just trying to give you a feeling of what happens to the object as it is tossed horizontally and attracted by the earth. Yes, $\Delta s$ represents an infinitesimal displacement as a result of the object's velocity, while $\Delta r$ is the simulataneous displacement as a result of the attraction of the earth. The combined displacement forms the orbit of the object. From the diagram you can see that if these two displacements are in a particular ratio then the velocity of the object will stay perpendicular to the attractive force and the speed of the object will stay the same, only the direction of the velocity will change.

Eliptical orbits it is easier to understand from an energy viewpoint. The potential energy is usually chosen to be zero at infinity. The object, m, start out with some total energy E (kinetic K, plus potential U), as it moves away from the force centre, it's potential energy increases at the expense of it's kinetic energy. If it's initial energy E < 0 it will stay bound to the force centre and it will eventually fall back to it thereby following an elliptical path with a play between K and U energy.

Last edited: Nov 7, 2005
11. Nov 8, 2005

### andrevdh

The shape of the orbit is determined by the (horizontal) launch velocity $v_o$ at a radius $r=c$ from the force centre. If $\mu=GM$ the one would get an elliptic orbit if
$${v^2}_o \ < \ \frac{2\mu}{c}$$. That is the only condition for an elliptical orbit to form.

Last edited: Nov 8, 2005