Energy involving satellite of the Earth

Homework Statement

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?

Homework Equations

PEg = -Gm1m2/R
Vescape = root(2)(G)(m1)/Rm
E binding = 1/2<PEg> in orbit[/B]

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

ehild
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Homework Statement

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?

Homework Equations

PEg = -Gm1m2/R
Vescape = root(2)(G)(m1)/Rm
E binding = 1/2<PEg> in orbit[/B]

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

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.

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.
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

ehild
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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

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.

• Glenboro
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.
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

haruspex
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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
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.

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.
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

ehild
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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
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.

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.
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 :(

haruspex
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a)PEg = -(6.67x10^-11)(5.98x10^24)(1550)/(7.00 x 10^6)
the gravitational potential energy of the satellite with respect to Earth

My small mistake, Thanks haruspex

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

SammyS
Staff Emeritus
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Can you give me a little hint :( My answers keep getting off
Did you recalculate the gravitational potential energy?

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

Did you recalculate the gravitational potential energy?

That velocity you have for part (b) looks more like orbital velocity than escape velocity.
Since it is in orbit, yes I calculated part (b) as orbital velocity.

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

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

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

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?
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)

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

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

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.
Would you mind if you can briefly explain how "relative" thing works? I need to put those into my head first SammyS
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Would you mind if you can briefly explain how "relative" thing works? I need to put those into my head first Another way to state the question is as follows:

What is the gravitational potential energy of the satellite in orbit compared to its gravitational potential energy when it was on Earth?​

By how much does it change and does it gain or does it lose potential energy?

Another way to state the question is as follows:

What is the gravitational potential energy of the satellite in orbit compared to its gravitational potential energy when it was on Earth?​

By how much does it change and does it gain or does it lose potential energy?
That is much better, thanks man