# Power and Satellites

1. Apr 7, 2006

### xc630

Hello I need osme help with these two problems. I was told the ansers but do not undersatnd how the answers were gotten.

1. The power dissipated in lifting a 2.0 kg ball vertically at 3.0m/s equals what?
The answer is 59 W. ALl I know is that power equals Work/ delta T but there is no time for this one.

2. Increasing the orbital radius of a satellite by a factor of 2 requires its period to change by a factor of what?

For thi I know R^3/T2=R^4/T^2.

I would appreciate any help.

2. Apr 7, 2006

### Staff: Mentor

In the first problem:

Work = Force * distance (and actually it is an itegral) but let's assume the force is constant.

Power = d/dt (Work) or d/dt (Energy).

Now, if a mass m, which is subjected to gravity of acceleration g, is moved up a distance, h, it experiences a change in potential energy of mgh. Now if it moves at constant velocity, v, then v = h/t, where t is the time it takes to move a distance h.

In the second problem, be sure the formula is correct:

See - http://en.wikipedia.org/wiki/Orbital_period#Calculation

http://en.wikipedia.org/wiki/Kepler's_laws_of_planetary_motion

3. Apr 7, 2006

### BobG

Your second one is wrong.

Kepler's Third Law: The square of the period is proportional to the cube of the mean distance.

In other words:
$$t = C a^{\frac{3}{2}}$$

If the radius is doubled, then you have:

$$t = C (2a)^{\frac{3}{2}}$$

C is just some constant, since Kepler's law just explains the relationship between the period and the average radius. However, C is equal to:
$$C = \frac{2 \pi}{\sqrt{G M}}$$
where G is the universal gravitational constant and M is the mass of the object the satellite is orbiting.

For Earth:
$$G M = 398600.4418 \frac{km^3}{sec^2}$$

If your satellite's radius was 42,164 km, your orbital period would be:

$$t=\frac{2 \pi}{\sqrt{398600.4418 \frac{km^3}{sec^2}}} * (42164 km)^{\frac{3}{2}}$$
which, coincidentally, happens to equal the number of seconds required for the Earth to rotate 360 degrees.

4. Apr 7, 2006

### Kyon

Hello,

The answer to the first question is much more simple than what Astronuc said. (He is not wrong at all though, but he uses big artillery for a really simple problem in my opinion ).

The power dissipated in lifting a 2.0 kg ball vertically at 3.0m/s equals what?
Here is my answer:

In this exercice, the only force that apply on the ball is the weight P=mg.
We know that: Power = Work / Time = (Force * Distance) / Time

With the infos provided, the simplest thing is to take 1 second for time:
Power = (2 * 9.81 * 3) / 1 = 58.86 # 59 W

You could have taken any amount of time, provided you calculate the associated distance.

What astronuc said is more of the general case, but is still totally true though.

I hope I could help,

Cyril

PS: Sorry if there are some english mistakes, I'm not a native english speaker ^^

5. Apr 8, 2006

### topsquark

Or, if the force and velocity are constant, P=Fv. But, of course, that was his end result anyway.

-Dan

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