# Work: Change in energy

1. Mar 6, 2015

### henry3369

1. The problem statement, all variables and given/known data
You wish to put a 1000-kg satellite into a circular orbit 300 km above the earth’s surface.
How much work must be done to the satellite to put it in orbit?

2. Relevant equations
Conservation of energy

3. The attempt at a solution
I don't need help answering the question because this is solved in my book. I'm confused about the calculation of work. My book solves it by using the change in total mechanical energy. Earlier in my book, it defined work as a change in kinetic energy or a negative change in potential energy. Why is it now a change in total mechanical energy?

2. Mar 6, 2015

### cajamarcus

I don't know the exact English terms, but I will try to explain what I know:

E: Energy [J]
P: Static pressure [Pa]
u: Velocity of the object [m/s]
G: Gravitational acceleration [m/s 2 ]
z: Elevation [m]
m: Mass of the object [kg]
V: Volume of the object [m3]
1: point 1
2: point 2

The Conservation of Energy:
Epressure+Ekinetic+Eelevation must be constant! (In ideal case. In reality there will also be Energy Losses)

P1*V+1/2*m*u12+m*g*z1=P2*V+1/2*m*u22+m*g*z2 + Loss

Basically, if the satellite goes up, that means it's elevation (potential) will increase; so you'll need to some kinetic or pressure energy to do it. I am not sure, how complex the question is. I mean, you can ignore the change in gravitational acceleration, and/or change in static pressure of the air; or you can also take them into account. If you ignore the both, then simply you can say:

In order to put the satellite into an orbit, you need to increase it's potential energy (elevation):
height = z2-z1 [m]
Eelevation=m*g*(z2-z1)
Assuming the air pressure is same in both elevations, you need only kinetic energy to put it into an orbit. In other words, kinetic energy will be transformmed to potential energy.
Ekinetic=Epotential
You can also calculate the required initial velocity by the folowing equation:
u1: The required velocity at the starting point [m/s]
u2: Since the satellite will no longer move after reaching to its orbit, u2 will be zero
Ekinetic=1/2*m*(u12-u22)
Ekinetic=1/2*m*u12

u=√(Ekinetic*2/m)

Last edited: Mar 6, 2015
3. Mar 6, 2015

### AlephNumbers

Caja, I don't think that trying to do the problem for Henry leads him to a better understanding of physics...

4. Mar 6, 2015

### cajamarcus

Sorry, you may be right; I will erase the calculation part.

5. Mar 6, 2015

### AlephNumbers

I don't think your solution is correct. The satellite would have to be moving once it reaches its orbit. If only enough work was done to the satellite to lift it to that elevation, then it would simply fall back to the earth.

6. Mar 6, 2015

### AlephNumbers

Henry, I think it is helpful to split this problem into two parts. First, how much work would need to be done on the satellite to lift it 300km above the earth?

7. Mar 6, 2015

### cajamarcus

I didn't consider the space effects actually. I assumed that it is a simple work-energy problem; so ignored many things. I assumed that once it reaches to the orbit, the net force will be zero, so the satellite will neither go further, nor fall back.

8. Mar 6, 2015

### AlephNumbers

That is not what I am saying. I would not be capable of solving this problem if it was anything but a "simple" work-energy problem. What I am saying is that in order for the satellite to orbit the earth, it must have a certain velocity. Otherwise it would fall back down to the earth. Work must to be done to get the satellite up, and work must also be done to keep the satellite up in orbit. If this is not apparent to you I would recommend reviewing potential energy, conservation of mechanical energy, gravity, centripetal force, and circular motion.

9. Mar 7, 2015

### henry3369

Sorry for the late response. I already know how to solve it. I use the change in total mechanical energy = work. I'm asking why its the change in total mechanical energy. Previously in my book, I was told that work is either the change in kinetic energy or the negative change in potential energy.

10. Mar 7, 2015

### henry3369

And using a change in potential or change in kinetic energy will yield incorrect results. Instead I have to use the change in total mechanical energy.

11. Mar 7, 2015

### BvU

Caja: welcome to PF ! You'll do fine and helping out on PF is a good steep towards becoming a real scientist !
Aleph: love your profile page !

Henry: Energy has lots of forms. In a mechanics course we focus on kinetic and potential energy (from springs or from gravity). Both forms of mechanical energy. Other forms of energy are heat, chemical energy, electrical energy, radioactive energy (that one's a bit complicated).

Energy is conserved (with a small reservation in this radioactive business; let's leave that aside for now).
But energy can be converted from one form into another. So if a car brakes, kinetic energy is converted into heat. Generally, friction converts kinetic energy into heat. Since potential energy and kinetic energy can be easily converted into one another, we take them together as mechanical energy. Think of a pendulum or an oscillating mass suspended from a spring.

You may want to check back in the book text if they really formulate it as you say. I would think that if I lift a book from the floor to a shelf, then the (positive) work I do adds a (positive) amount of potential energy to the book.

Work is force times displacement in the direction of the force. A so-called vector product: $W = \vec F\cdot\vec {\Delta x}$. So for the book I did $mg\Delta h$ of work and its potential energy increased by $mg\Delta h$
The book on the shelf can drop off and "convert the work I did previously" into kinetic energy: ${1\over 2} mv^2 = mg\Delta h$. When it lands, all that is converted into other forms of energy (noise, heat, you can pulverize a pack of potato chips, etc.).

All in all it's not unreasonable to lump kinetic energy and potential energy together when doing this kind of mechanics problems.

I could go on for pages about this energy (heat is also mainly kinetic energy, but on an atomic scale -- later).

And the radioactive stuff brings in mass too. Now we're already at E = mc2 and that's a bit too much at this stage. It'll come later.

12. Mar 7, 2015

### Staff: Mentor

The total mechanical energy for an object in an isolated gravitational system is a constant, and in fact is a parameter that can be used to determine several facts about the orbit of the object such as whether or not the object is gravitationally bound to the system and if so, the size of the orbit (radius if circular, major axis if elliptical). So it's a handy concept.

When work is done that changes the total mechanical energy (by external forces or by tapping other forms of energy such as chemical rockets) then the size of the orbit changes.

In this problem, in order to bring the system to the desired configuration you need to do work to change the potential energy and work to change the kinetic energy. That is, you need to change both components of the total mechanical energy.

You can consider the work done to change the KE and PE separately if you wish, then sum them to arrive at the total work required. If you consider the form of the total mechanical energy though, you will see that this is just the resulting change in total mechanical energy.

13. Mar 7, 2015

### PhanthomJay

You likely misinterpreted the book definition of work. There is work done by conservative forces and work done by non conservative forces. Work done by conservative forces like gravity is the negative of the change in potential energy. Work done by non conservative forces like rocket thrust is the (change in kinetic plus change in potential energy) that is, the change in mechanical energy. The sum total of these two is the change in kinetic energy

Last edited: Mar 7, 2015