Energy involving satellite of the Earth

1. May 8, 2015

Glenboro

1. The problem statement, all variables and given/known data
A satellite of the Earth has a mass of 1550 kg. It orbits Earth with a mean radius of orbit of 7.00 x 10^6 m.

a) What is the gravitational potential energy of the satellite with respect to Earth?

b) What is the kinetic energy and the velocity of the satellite in Earth’s orbit?

c) What is the binding energy of the satellite to Earth?

d) What is the total mechanical energy of the satellite in its orbit?

2. Relevant equations
PEg = -Gm1m2/R
Vescape = root(2)(G)(m1)/Rm
E binding = 1/2<PEg> in orbit

3. The attempt at a solution

a)PEg = -(6.67 X 10^-11)(5.98 X 10^24)(1550)/(1.338 X 10^7)
= -4.62 X 10^10J

b) Vescape = root(2)(6.67 x 10^-11)(5.98 x 10^24)/(6.38 x 10^6)
= 1.12 x 10^4 m/s
KE = (1/2)(m)(V)^2
KE = (1/2)(1550)(1.12 x 10^4)^2
KE = 9.72 x 10^10J
c) E binding = 1/2<PEg> in orbit
(0.5)(4.62 X 10^10J) = 2.31 x 10^10J

d) I'm not certain which equation I should use for this one

2. May 8, 2015

ehild

The total mechanical energy consist of the kinetic energy + the potential energy.
You calculated the kinetic energy form the escape velocity. That is wrong. The satellite is on orbit, so its speed is less.

3. May 8, 2015

Glenboro

So for question d) I will just use equation Me = Ke + Pe
and for b) Ke = 1/2<PEg> which is 2.31 x 10^10J or I might be completely off

4. May 8, 2015

ehild

KE is not half of the PE. They do not have the same sign.
I do not understand what radius did you use when calculating the PE.

5. May 8, 2015

Glenboro

The radius I used to calculated was (6.38 x 10^6m) The earth's radius plus 7.00 x 10^6 m from the question. (7.00 x 10^6m)+(6.38 x 10^6) = 1.338 x 10^7 m
Is this what you are referring to? Sorry for making mistake, I actually trying my best

6. May 8, 2015

haruspex

The 7 106 is not the height above the Earth's surface, it's the radius of orbit, so is measured from the Earth's centre.

7. May 8, 2015

Glenboro

Oh so to figure out how far it's distance from earth I had to subtract (6.38 x 10^6m) from 7.00 x 10^6

8. May 8, 2015

ehild

The distance from the Earth surface is irrelevant. The satellite orbits Earth with a mean radius of orbit of 7.00 x 10^6 m. The force between the Earth and satellite and the potential energy depend on the distance from the Earth centre.

9. May 13, 2015

Glenboro

So I fully realized my error, I should known it was the radius of orbit. I did re-calculation for all questions.
a)PEg = -(6.67x10^-11)(5.98x10^24)(1550)/(7.00 x 10^6)
= -8.83 x10^10J
b) I didn't had to use Vescape formula since it was in orbit
V= root (G)(Me)/R
v= root (6.67x10^-11)(5.98 x10^24)/(7.00 x 10^6)
V = 7.55 x 10^3 m/s
Ke = (0.5)(1550)(7.55 x 10^3)^2
Ke = 4.42 x 10^10J

If this isn't right, I'm dumb :(

10. May 13, 2015

haruspex

11. May 13, 2015

Glenboro

My small mistake, Thanks haruspex

12. May 13, 2015

Glenboro

Can you give me a little hint :( My answers keep getting off

13. May 13, 2015

SammyS

Staff Emeritus
Did you recalculate the gravitational potential energy?

That velocity you have for part (b) looks more like orbital velocity than escape velocity.

14. May 13, 2015

Glenboro

Since it is in orbit, yes I calculated part (b) as orbital velocity.

15. May 13, 2015

SammyS

Staff Emeritus
Oh ! yes, of course, that's right.

Have you corrected you answer to part (a) so the answer is relative to Earth?

What was the satellite's gravitational potential energy when it sat on Earth's surface?

16. May 13, 2015

Glenboro

Gravitational potential energy?
PEg = mgh
PEg = (1550kg)(7.00 x 10^6m)(g)

17. May 13, 2015

SammyS

Staff Emeritus
That's good for objects very near earth's surface, and is relative to being exactly at earth's surface, where h = 0.

At earth's surface, how far is an object from the center of the earth?

18. May 13, 2015

Glenboro

Yeah because gravity would not affect much when it's it orbit. Due to distance
At Earth's surface, how far is an object from the centre of the earth?
(7.00 x 10^6m) - (Earth's radius)

19. May 13, 2015

SammyS

Staff Emeritus
Let's review part (a) of your question.

You have found the gravitational potential energy of the satellite in its orbital position, but that is not with respect to earth. (What is that with respect to?)

What was the satellite's gravitational potential energy when it sat on Earth's surface? That's not in orbit.

20. May 13, 2015

Glenboro

Would you mind if you can briefly explain how "relative" thing works? I need to put those into my head first