Minimum Velocity for Satellite Escape from Earth's Attraction

In summary, Joe calculated that the minimum velocity that the satellite meteosat (time period is 23h56min and altitude is 35800km + radius of earth(6370km)) must be given to escape from the force of attraction of the Earth is 3078ms^-1. Daniel calculated that the potential energy of the satellite is -GMm/2R. Together, they figured out that the escape velocity is 4349 m/sec.
  • #1
josephcollins
59
0
Hi ppl, could someone just clarify this for me

i need to find the minimum velocity that the satellite meteosat (time period is 23h56min and altitude is 35800km + radius of earth(6370km)) must be given to escape from the force of attraction of the Earth.

I started by saying that the total energy needed to effect this is GMm/2(R) where M is mass of earth, m is mass of satellite and R is orbit. Then I equated this to mv^2/2

My final answer was 3078ms^-1 Could someone verify my working or at least show me where I made an error if there is one?

Thanks, Joe
 
Physics news on Phys.org
  • #2
Yes,your answer is correct,as well as the method to get it...

Daniel.
 
  • #3
josephcollins said:
Hi ppl, could someone just clarify this for me

i need to find the minimum velocity that the satellite meteosat (time period is 23h56min and altitude is 35800km + radius of earth(6370km)) must be given to escape from the force of attraction of the Earth.

I started by saying that the total energy needed to effect this is GMm/2(R) where M is mass of earth, m is mass of satellite and R is orbit. Then I equated this to mv^2/2
Why 2(R) in the denominator? I think it should be:

[tex]\frac{1}{2}v^2 = \frac{GM}{R}[/tex]

[tex]v = \sqrt{2GM/R}[/tex]

[tex]v = (2*6.67e(-11)*5.98e24/4.217e7)^.5 = 4349 m/sec.[/tex]

AM
 
  • #4
Andrew Mason said:
Why 2(R) in the denominator? I think it should be:

[tex]\frac{1}{2}v^2 = \frac{GM}{R}[/tex]

[tex]v = \sqrt{2GM/R}[/tex]

[tex]v = (2*6.67e(-11)*5.98e24/4.217e7)^.5 = 4349 m/sec.[/tex]

AM

I think so too.
 
  • #5
The reason I took 2r as the denominator is that GMm/2r is the total energy of the satellite at that position in orbit, whereas GMm/r is just the potential...

Is this correct?
 
  • #6
Yes,my guess is that u were right in the first place,as the satellite certainly has KE,not only PE.If my memory doesn't cheat me,it is your formula...

However,i may be wrong...

We need some other opinion.

Daniel.
 
  • #7
josephcollins said:
The reason I took 2r as the denominator is that GMm/2r is the total energy of the satellite at that position in orbit, whereas GMm/r is just the potential...

Is this correct?

Can you explain why you think the satelite's energy is -GMm/2r ? Does this include it's kinetic energy?
 
  • #8
Yes.It's the only possible reason.

Daniel.
 
  • #9
Assuming circular motin the satelites potential energy is:
[tex]-G\frac{Mm}{r}[/tex]
It's kinetic energy is:
[tex]\frac{1}{2}m\frac{(2 \pi r)^2}{T^2}[/tex].

Keplers law gives

[tex]T^2=\frac{(2 \pi r)^2 r}{GM}[/tex]

so the kinetic energy can be written
[tex]G \frac{Mm}{2r}[/tex].
The total energy is kinetic plus potential:

[tex]-G\frac{Mm}{2r}[/tex], so you're correct.

At infinity it's total energy wil be zero, so you'll have to add:
[tex]\frac{1}{2}mv^2 = G\frac{Mm}{2r}[/tex].
So the escape velocity is:

[tex]v_{esc}=\sqrt{G\frac{M}{r}}=3075ms^{-1}[/tex]

So you're totally right.

EDIT: As Andrew Mason pointed out, you (and I) calculated the extra velocity required to escape, not the escape velocity...
 
Last edited:
  • #10
josephcollins said:
The reason I took 2r as the denominator is that GMm/2r is the total energy of the satellite at that position in orbit, whereas GMm/r is just the potential...

Is this correct?
Not quite. It is -GMm/r. It is important to include the - sign in the potential.

The question asks for the minimum velocity needed to escape so it doesn't matter what velocity it has while orbiting. But if you were asked what the additional speed is required, it would be the difference between the escape velocity and the orbital velocity. You calculated the orbital velocity, not the escape velocity.

ie. The energy of the satellite in orbit is:

(1)[tex]E_{orbit} = KE + PE = \frac{1}{2}mv^2 - \frac{GMm}{r}[/tex]

Since:
[tex]\frac{mv^2}{r} = F = \frac{GMm}{r^2}[/tex]

(1) reduces to:

[tex]E_{orbit} = \frac{1}{2}\frac{GMm}{r} - \frac{GMm}{r}[/tex]

[tex]E_{orbit} = - \frac{1}{2}\frac{GMm}{r}[/tex]

Since the condition for escape is that KE + PE = 0, the kinetic energy needed for escape is twice the orbital kinetic energy. This requires increasing the kinetic energy by factor of 2 or its speed by [itex]\sqrt{2}[/itex]

AM
 
  • #11
Since the question asks for the escape velocity rather than the root sqaure excess velocity, you must use R and not 2R. I concur with AM.

[tex] KE(in) + PE(in) = E(fin) = 0 [/tex]

[tex]\frac{1}{2}mv_{esc}^2 - \frac{GMm}{R} = 0 [/tex]

That's all there is to it.
If you add the existing KE to the PE to get _GMm/2R, you are then only finding the extra KE required. This is not what's asked.

da_willem said:
At infinity it's total energy wil be zero, so you'll have to add:
[tex]\frac{1}{2}mv^2 = G\frac{Mm}{2r}[/tex]
So the escape velocity is:
[tex]\frac{1}{2}mv^2 = G\frac{Mm}{2r}[/tex]
How did you go from additional KE to escape velocity ?
 
  • #12
Okay,Gokul,which part didn't u understand??This one:

josephcollins said:
i need to find the minimum velocity that the satellite meteosat (time period is 23h56min and altitude is 35800km + radius of earth(6370km)) must be given to escape from the force of attraction of the Earth.

Please explain to me how would you keep a geostationary satellite in orbit 42000Km (from the Earth's center) MOTIONLESS... :uhh:

Daniel.
 
  • #13
Gokul43201 said:
How did you go from additional KE to escape velocity ?

Read the EDIT one line below...
 
  • #14
da_willem said:
EDIT: As Andrew Mason pointed out, you (and I) calculated the extra velocity required to escape, not the escape velocity...
Not exactly. The additional velocity is [itex](\sqrt{2}-1) \times 3075 = 1274 m/s.[/itex]

AM
 
  • #15
dextercioby said:
Please explain to me how would you keep a geostationary satellite in orbit 42000Km (from the Earth's center) MOTIONLESS.
No one said it is motionless. Its orbital speed is 3075 m/s. It needs to have a speed of 4349 m/s (or twice the kinetic energy than 3075 m/s) in order to escape.

AM
 
  • #16
dexter, if you see a mistake in my post, point it out. And take it easy with the large fonts !

Your calculation does not even find the correct additional velocity required, as AM pointed out above. It only calculates what I refer to as the "root excess square velocity", which is quite a meaningless number for the given problem.
 

1. What is the definition of minimum velocity?

Minimum velocity refers to the lowest velocity at which an object can move or the minimum speed required to start a particular motion.

2. How is minimum velocity calculated?

Minimum velocity is calculated using the formula v = √(2gh), where v is the minimum velocity, g is the acceleration due to gravity, and h is the height of the object.

3. Why is it important to find the minimum velocity?

Finding the minimum velocity is important in various fields of science, such as physics and engineering, as it helps determine the minimum energy required for an object to move or the minimum force needed to overcome friction.

4. How does air resistance affect the minimum velocity?

Air resistance can increase the minimum velocity needed for an object to move due to the opposing force it creates, making it harder for the object to overcome its initial inertia.

5. Can the minimum velocity be greater than the escape velocity?

No, the escape velocity is the minimum velocity required for an object to escape the gravitational pull of a planet or celestial body. Therefore, the minimum velocity cannot be greater than the escape velocity.

Similar threads

Replies
14
Views
1K
Replies
6
Views
2K
Replies
13
Views
8K
Replies
12
Views
4K
Replies
2
Views
933
Replies
43
Views
5K
Replies
7
Views
2K
Replies
3
Views
3K
  • Introductory Physics Homework Help
Replies
7
Views
1K
  • Introductory Physics Homework Help
2
Replies
39
Views
3K
Back
Top