# Quick tangential speed question

• jj8890
You should getv \sim \frac 1 rThat is, the orbital velocity is inversely proportional to the radius of the orbit. For the B satellite, that radius is 4 times as large, so the orbital velocity is (1/4) times the orbital velocity of the A satellite. In summary, the ratio of the tangential speed of satellite B to that of satellite A is 1/4.
jj8890

## Homework Statement

Two satellites A and B of the same mass are going around the Earth in concentric orbits. The distance of satellite B from the earth’s center is four times that of satellite A. What is the ratio of the tangential speed of satellite B to that of satellite A?

v=rw

## The Attempt at a Solution

So if v=rw and B is 4 times the distance of A, wouldn't the ratio (V_B/V_A) just be:
4rw = rw
=4

Think again, this time using Kepler's laws.

I'm sorr, i don't see wher eyou are coming from.

Kepler's Laws are:
1. The path of the planets about the sun are elliptical in shape, with the center of the sun being located at one focus. (The Law of Ellipses)
2. An imaginary line drawn from the center of the sun to the center of the planet will sweep out equal areas in equal intervals of time. (The Law of Equal Areas)
3. The ratio of the squares of the periods of any two planets is equal to the ratio of the cubes of their average distances from the sun. (The Law of Harmonies)

But I don't see how these apply if I do not have a period and am only trying to calculate the ratio? Doesn't v=rw? Or am i totally off base?

jj8890 said:
3. The ratio of the squares of the periods of any two planets is equal to the ratio of the cubes of their average distances from the sun. (The Law of Harmonies)

But I don't see how these apply if I do not have a period and am only trying to calculate the ratio? Doesn't v=rw? Or am i totally off base?

The satellites are in orbit. Of course they have a period, and because the satellites are at different orbital altitudes their periods are not the same per Kepler's third law. You don't know exactly what the period is, but that doesn't matter here. You only need to know the ratio of the periods.

What is the relationship between orbital period T and orbital angular velocity w?

The angular velocity specifies the angular speed at which an object is rotating along with the direction in which it is rotating so it is a portion so...when i looked it up the period equals (w/2*pi)^-1 where w is the angular velocity...where would I go from there?

jj8890 said:
The angular velocity specifies the angular speed at which an object is rotating along with the direction in which it is rotating so it is a portion so...when i looked it up the period equals (w/2*pi)^-1 where w is the angular velocity...where would I go from there?
You wrote Kepler's third law in English. Write it as a mathematical equation instead. Next, write an alternative version of the law that uses angular velocity in lieu of period. Finally, apply this to the problem at hand.

T^2=r^3 so since t=w/2*pi^-1:
r^3=(w/2*pi^-1)^2

I'm sorry but i feel so confused I appreciate you being patient with me but where would I go from here. The thing is that we haven't used angular velocity in my class yet and i didn;t know that the period equaled w/2*pi^-1 until i looked about 3 chapters ahead. So, does this look right? How would I apply this to the problem at hand?

First, you can get rid of the pi. Kepler's third law is a proportionality, not an equality.

Second, if you haven't learned about angular velocity yet, why did you bring it up in your original post? All you need is Kepler's third law and the formula for the circumference of a circle. You know the formulation; you even used it in a different thread, quoted here:

jj8890 said:

## Homework Equations

v= (2*pi*R)/T

So, if v=2*pi*R/T and T^2=r^3, wouldn't v/2*pi*r =r^3 and one r would cancel out making it v=2*pi*r^2 so for the ratio, 2*pi *(4)r^2 =2*pi * r^2 and everything would cancel out except the 4 making the answer 4 like I originally thought?

No. Think about it this way. Saturn is about 10 AU from the Sun (the Earth is 1 AU). By your logic, Saturn should have about 10 times the Earth's orbital velocity. However, Saturn has a period of about 30 years. Coupling that with the distance means that has about 1/3 of the Earth's orbital velocity, not 10 times the Earth's velocity.

So the ratio would be 1/4 or would you have to square the r term making it 1/16?

No. Look at Saturn again. The ratio is not 1/100, or even 1/10. It is about 1/3.

Kepler's third law:

$$T^2 \sim r^3$$

Note that I did not use the equal sign in the above expression. Kepler said the square of the period is proportional to the cube of the average distance. This is a proportionality, not an equality. Taking the square root of both sides,

$T \sim r^{3/2}$

You already know $v=2\pi r/T$, or, as a proportionality,

$$v \sim \frac r T$$

You have an expression for T in terms of r. Apply that to the above.

## 1. What is tangential speed?

Tangential speed is the linear speed of an object as it moves along a circular path.

## 2. How is tangential speed different from angular speed?

Tangential speed is the linear speed along the circular path, while angular speed is the rate of change of the angle of rotation.

## 3. What is the formula for calculating tangential speed?

The formula for tangential speed is v = rω, where v is tangential speed, r is the radius of the circular path, and ω is the angular speed.

## 4. Can you provide an example of tangential speed calculation?

For example, if a car is traveling on a circular track with a radius of 10 meters and an angular speed of 2 radians per second, the tangential speed would be 20 meters per second (v = 10 meters * 2 radians/second).

## 5. How is tangential speed important in science?

Tangential speed is important in various scientific fields, such as physics, engineering, and astronomy. It helps us understand the motion of objects along circular paths and is essential for calculations involving circular motion and rotational dynamics.

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