What am I calculating here?

[SpaceTime]

If I divide a planets orbit velocity by the planets orbit radial distance from the sun, what is the resulting unit definition (frequency? or angular velocity?) and how are the actual units defined or clarified in value?

 

Signed

SpaceTime

[Dapthar]

If I divide a planets orbit velocity by the planets orbit radial distance from the sun' date=' what is the resulting unit definition (frequency? or angular velocity?) and how are the actual units defined or clarified in value?

[/quote']Well, orbital velocity has units of [math]\frac{meters}{second}[/math], and radial distance has units of [math]meters[/math], so dividing these two quantities, we get [math]\frac{\frac{m}{s}}{m} = \frac{1}{s} = Hertz[/math].

 

As to the physical significance, I suppose it would be the frequency of the planet's revolution if it were confined to a circle whose circumference was the planet's original radial distance from the aforementioned star.

 

However, I believe that you meant to ask for the frequency of the planet's revolution about a star, which would be the planet's orbital velocity divided by the distance it travels in one revolution, which is equal to [math]2\pi r[/math], where [math]r[/math] is the planet's radial distance from its star in meters.

[SpaceTime]

Hi Dapthar,

Thanks for the quick response. The information you supplied is not nescessary the answer I was looking for.

 

If I divide the orbital circumference with the orbital velocity then this would give the frequency in Hertz?

 

But the two vectors of the orbital velocity and the radius length are 90 degrees out of phase, surely then the answer is eqivalent to Hertz / 2pi ?

 

From this perpective what am I actually calculating? is it angular velocity for the frequency?

 

Signed

SpaceTime

[The Rebel]

In short, it is the mean angular velocity.

 

I say mean because most orbits are elliptical such that the velocity changes depending on the radius at each point.

[SpaceTime]

Hi Rebel,

Thanks for the definate clarification. Could you extend to me one more courtesy, or anyone else who may know.

 

If we used the previous example as the mean angular velocity for earth in its orbit around the sun and then wanted to include the earths daily spin angular velocity

( i.e. spin velocity / equatorial radius = 462.9 kms / 6378000 km ) so as to enable us to obtain a combined mean resultant angular velocity value.

 

What would you suggest as being the correct approach, is it as simple as:

 

Resutant angular velocity = (earth angular velocity / solar angular velocity)

 

And would the answer be, in meters per second or hertz?

 

Thanks in advance

SpaceTime

[mezarashi]

To the original question:

 

If you divide the linear velocity (m/s) by the orbital radius (m), then you will get the angular velocity in radians per second. Frequency would refer to revolutions per second. Doing a simple calculation using for example a 10m/s linear velocity and a 10m radius will verify that this is correct.

 

To SpaceTime:

 

Well first off I'll need to assume an approximate circular orbit. I don't have experience with calculations involving elliptical myself. In anycase, the correct approach I think would be an addition of the resultant vectors.

 

This concept is more clear with the use of vectors. The angular velocity vector actually points in the direction perpendicular to both the radius vector and the instantaneous linear velocity vector. The velocity relationship: v(linear) = OMEGA(angular) x r(radius) as found in AP Physics or more advanced dynamics courses.

 

In effect, if the Earth is circling the sun in a counter clockwise fashion, then this vector points up (by the right hand rule). Now you can also similarly consider the angular velocity of the Earth on its axis, and add the parallel vector components. The Earth is tilted by 23 degrees or so I believe. The answer is in radians per second (the standard unit for angular calculations).

 

Edit: All the radians -> radians/sec

[SpaceTime]

I am not absolutely clear on this, we for simplicity sake will assume circular orbit and rotation and ignore the fact that the earths tilts.

 

The rotating earth is travelling at an orbital velocity of 29770 metres per second and sinultaneously has a equatorial surface spin of 463 meters per second.

 

How would I calculate the combined resultant equatorial value? could you supply an example for me, both for forward and reverse equatorial spin.

 

Thanks again.

SpaceTime

[mezarashi]

Hi there, sorry, I didn't follow up on this post. Let me draw a picture for you to illustrate.

 

Update:

 

 

Don't ask me why the vector math is this way. Ask the mathematicians. They've given us engineers the tools that are consistent for the calculations needed.

[SpaceTime]

Maybe I still dont understand.

 

Sun radius is 149,600,000,000 meters

Orbit velocity 29,770 meters sec

 

Orbit Angular velocity = 29,770 / 149,600,000,000 = 1.98997E-7

 

Earth spin angular velocity = 462.95 / 6,378,000 = 7.258431E-5

 

How would I mathematically obtain the combined result of these two conditions under these conditions.

 

Is one divided by the other i.e.

 

result = (1.98997326203209E-7 / 7.2584311942428E-5) = 0.00274

 

Is this the correct way to define this point on the earths surface?

 

Signed

SpaceTime