speed of sphere?

Hi there,

I don`t do much mechanics or physics, so i was wondering how you work this out:

If you knew the speed of rotation of (any point on the equator of) a sphere, and if you also knew the speed of that sphere in a circular orbit about another, bigger sphere, how would you work out the TOTAL speed of a point on the equator of the smaller sphere - since it`s revolving and moving at the same time?

I don`t know how to work that out - and i`m just curious because i think it`s an interesting question!

Reply 1

kingkongjaffa

Original post by DrSheldonCooper

It doesn't have a compounded total speed it has two speeds based on the two axis' of rotation, the rotational speed about the axis of the small ball, and the rotational speed of the ball about the axis of the larger ball it's orbiting.

(edited 11 years ago)

Reply 2

Hopple

Original post by DrSheldonCooper

I'd say you just add them together (or subtract, depending if the orbit and rotation are in the same direction).

Original post by kingkongjaffa

A point on the equator would have a speed that you could measure :s-smilie:

Reply 3

kingkongjaffa

Original post by Hopple

I'd say you just add them together (or subtract, depending if the orbit and rotation are in the same direction).

A point on the equator would have a speed that you could measure :s-smilie:

yeah of course it would, it's rotational speed about the spheres own axis :s

but orbit and rotation are two seperate velocities, theres no reason for a common velocity you don't measure the moons rotational velocity around it's self and the obit of the earth and add them together do you. which is what OP wasrefering to.

Reply 4

User570431

Original post by DrSheldonCooper

To work out that point, you see the time and use the S=D/T formula :tongue:

Reply 5

Hopple

Original post by kingkongjaffa

Why don't you want there to be an answer? You know what the OP was asking for.

Possibly it can be used to work out the best speed of approach for landing on a planet/moon, you'd want the lander's approach to be roughly in sync with the planet's rotation.

Reply 6

kingkongjaffa

Original post by Hopple

because there isn't an answer it's two separate velocities what's so difficult about there's no resultant velocity due to the combination of them both.

And yes you would want the lander to be close to the orbital velocity so why do you need to know the velocity of the moon's orbit around the earth since you're both orbiting the earth at a similar radius that velocity is unimportant.

you work with one or the other you don't combine them both.

Reply 7

kingkongjaffa

Original post by blueray

To work out that point, you see the time and use the S=D/T formula :tongue:

no that only applies in linear motion with a constant acceleration.

the motion described here is circular or rotational, what distance would you measure, the only thing you can measure is the number of rotations around the point in radians or degrees turned through the axis.

Reply 8

Hopple

Original post by kingkongjaffa

Just because you don't come across what the OP asked every day doesn't mean it isn't an answer. The concept being discussed is a sane one, and there's no harm in quantifying it.

Reply 9

kingkongjaffa

Original post by Hopple

Just because you don't come across what the OP asked every day doesn't mean it isn't an answer. The concept being discussed is a sane one, and there's no harm in quantifying it.

but it's meaningless.

if you divide a constant velocity by time you get acceleration yes.

if you perform any operation on the two velocities you don't get any meaningful result.

it's like saying I'll times density by the speed of light because I want to

sure you can do it but it doesn't yield any meaningful.

Reply 10

Hopple

Original post by kingkongjaffa

I've just given you an example of how it has meaning. How fast would you need to get a spacecraft moving so that you could drop the lander? Possibly another use would be calculating red-shift, we'd need to know our relative speed to the Sun to work out the object's relative speed to the Sun - true our spinning velocity would be small compared to our orbitting velocity, but more accuracy can't hurt.

Reply 11

Mathlete18

Since the OP's question is interested only in the speed of such a point, surely one could find seperately the tangential speed and radial speed of the point at any given time, regarding the bigger sphere as the centre of rotation.

For example, let V = speed of smaller sphere along radial path
and let v = speed of rotation of the point on the smaller sphere

Surely the resultant tangential speed, Vt = V + vcos(wt + e)
and the radial speed, Vr = vsin(wt + e)

Where e = the phase shift, depending on where you start t0, and w = 2pi/T where T is the time period of the smaller sphere's rotation.

Thus the speed in question would = ((Vt)^2 + (Vr)^2)^.5

For which I get speed = (V^2 + 2vVcos(wt+e) + v^2)^.5

Hope this helps even though I may be completely off the mark here!

Reply 12

Manitude

I'd imagine the speed would be vary in some sinusoidal relation - the rotation of one point on the smaller planet is sometimes in the same direction as the motion of the planet and sometimes against it.

Reply 13

msmith2512

Original post by DrSheldonCooper

Suppose that the radius of the orbit is R and the small sphere is r. Suppose the angular velocity of the small sphere about the second sphere is w and the angular velocity of the small sphere about its axis is v.

Suppose the plane of the orbit is the xy-plane. Take a sensible start point then:

(x(t), y(t)) = (Rcoswt + rcosvt, Rsinwt + rcosvt)

This is the position vector of a point on the small sphere relative to the centre of the system. It's velocity (relative to this central point) will just be the time derivative of the position vector.