# Power and Satellites

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.

Astronuc
Staff Emeritus
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

BobG
Homework Helper
xc630 said:
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.

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

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

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?

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 ^^

Kyon said:
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?

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 ^^

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