# Orbital motions about the earth.

1. Jun 22, 2006

### al_201314

Hi all,

I've got some conceptual questions that I need to clear before my exam I hope someone can help me out here.

I understand that a geostationary satellite must orbit in the plane including the earth's centre, in order to stay above a particular surface of the earth.

However, can a satellite orbit in vertical circles (sorry I can't find a better term for this) i.e, from north pole all the way down to the south and back up to the north pole again? There is no mention of any satellites that does this in my book and it seems that they always put emphasis on the need for them to orbit in the plane including the earth's centre. Thus, can it also orbit, anywhere else around the earth?

Does anyone have the mathematical relation on g and the centripetal force at the equator of the earth? I was told that the centripetal force is less at the equator, since it must provide for the g?

But I couldn't get a worked out mathematical formula for this, say compared to at the poles. mv^2/r = Gmm/r^2. How does the g come in here?

Thanks a lot everyone.

Last edited: Jun 22, 2006
2. Jun 22, 2006

### al_201314

After some thinking through, I realised that one would weigh more at the equator, than at a pole, is this correct? My reasoning is that, assuming the earth is spherical, g would be the same at the pole as well as at the equator. But since the earth is spinning, centripetal acceleration acts on the person, which increases the net acceleration which would make him weigh more, is this right?

3. Jun 22, 2006

### tony873004

No. A person is lighter at the equator than at the poles. If the Earth spun fast enough you'd get flung off. So the faster it spins, the less you weigh.

Because of this, the Earth buldges at the equator. When you stand on the equator, you are farther from the center of the Earth than when you are standing on a pole. Therefore, by the equation g = GM/r^2, a slightly bigger r gives you a slightly smaller g.

A satellite can orbit in vertical circles. They're called polar orbits. Just watch an hour after sunset for satellites in the sky. You'll notice that quite a few of them, including the International Space Station move in North / South directions. The ISS isn't polar, but it is closer to polar than equatorial. This was done to make it easy for the Russians to launch to it.

A geosychronous satellite can be in a polar orbit, but a geostationary orbit cannot. As you said, geostationary orbit stays above the same particular point on Earth's surface. You can't do this with an orbit that visits both the North and the South pole. But if you orbit over the equator, and your orbital velocity is equal to the Earth's rotation velocity, then the Earth will spin with you and will remain above a fixed point on Earth's surface.

Satellites that orbit the Earth in 24 hours, but are not in equatorial orbits are said to be in geosychronous orbits. Your orbit's period and the period of Earth's rotation are sychronized. But you will wander North and South of the Equator, hence you will not be stationary over a single point.

A geostationary orbit is an ideal condition that is probably never achieved if you measure accurately enough. Even communications satellites will have a little slop in their orbits, making them technically geosychronous rather than geostationary. But the slop must be small enough that the satellite dishes installed on people's rooftops for TV don't need motors to track them.

Hope this helps.

4. Jun 22, 2006

### tony873004

Your book should have the rotational force and acceleation formulas. Its the same one used for merry-go-round problems, etc.

Combine this with the formula for acceleration due to gravity (after looking up the Earth's radius at the poles and the equator) and you will find out exactly how much g changes between the poles and the equator.

After computing it, do you think that it exceeds the margin of error of a typical bathroom scale?

5. Jun 22, 2006

### al_201314

As what I can understand from this, is it correct to say that when we rotate with the earth, some of the gravitational force is needed to keep us on the earth's surface. If this is all that is present, we would experience "weightlessness" because the normal force exerted by the ground on us is effectively = 0? And it is because gravitational forces are much more than just that, we would be pulled inwards in the ground was not there? And therefore the resultant force (weight) we experience is the gravitational force acting on us minus the amount that gives us the centripetal force? However, how does the gravitational force preovide with the centripetal force? (Assuming I'm reasoning correct above this)

From all this, assuming I am correct (kindly correct me if I'm wrong), is it the centrifugal force that makes us feel weightless?

Thanks for all the help and explanation.

Last edited: Jun 22, 2006
6. Jun 23, 2006

### tony873004

Yes, the centrifugal force makes us lighter. If Earth rotated at about 8 km/s at the equator, you would be weightless. This by no coincidence is orbital velocity for the surface of the Earth.

The formula for circular orbital velocity is Vc=sqr(GM/r). You'll find you get the same answer with v=sqr(ar), where a is acceleration at Earth's surface (9.8 m/s/s), r is the radius of Earth (6378000m), M is the mass of the Earth 5.97e+24kg, and G is the gravitational constant 6.672e-11.

7. Jun 23, 2006

### Hootenanny

Staff Emeritus
I am hesitant here with introducing centrifugal force into this concept; I would much prefer to discuss the apparent strength of gravity with respect to the centripetal force. The OP had it half right when he said that the centripetal force acts on the person, but he should have said it makes them feel less heavy.

Last edited: Jun 23, 2006
8. Jun 23, 2006

### al_201314

Thanks guys for the help.

Mind elaborating on this?

Sometimes I think the centrifugal/centripetal force concept is a bit confusing. We have a centripetal force acting on us say when we're in a bus turning right, directed towards the centre of the circle. On the other hand we feel we are being pushed out of the circle by the centrifugal force.

9. Jun 24, 2006

### Hootenanny

Staff Emeritus
Centrifugal force is a fictitious force, it only occurs if we view circular motion from a rotating reference frame. For example, if you are stood up on a playground roundabout which is spinning, you feel yourself being 'pulled' outwards by the 'centrifugal' force. This is because your reference frame is rotating, therefore it is accelerating and is a non inertial reference frame. However, if you are stood off the roundabout (inertial reference frame) watching someone else on a roundabout, you don't see any force 'pulling' them off, you only see them holding on to the bars, pulling towards the centre of rotation, this is the centripetal force.

Back to the example in hand. The centripetal force, isn't a force per se; it is simply the resultant force. For an object travelling in a uniform circle, the resultant force, must be directed towards the centre of the rotation, this resultant force is the centripetal force. Now, when you are stood on the earth there are two forces acting; gravity and the normal reaction force (N), they a directed in opposite directions, so we can form an expression;

$$F_{net} = mg - N = F_{centripetal}$$

The weight we feel is the normal reaction force of the earth, we feel weightless when N = 0;

$$N = mg - F_{centripetal} = mg - mr\omega^2$$

[tex]N = mg - mr\omega^2[/itex]

Note, that this equation only holds if we are at the equator or poles. Now, using the above equation it is easy to observe what will happen to the normal reaction force (our apparent weight) if we increase the angular frequency.

Last edited: Jun 24, 2006