Escape Velocity Calculation for an Object on Earth

In summary, the gravitational potential energy is equal to the gravitational potential energy of an object multiplied by the mass of the object. The kinetic energy of an object is equal to the product of its mass and the escape velocity.
  • #1
Stratosphere
373
0
How do you Calculating the escape velocity? I googled it but they give bad instructions on how to do it. What is the formula?
 
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  • #2
Every time I try using the formula i have i have to solve for V^2 and I get it to equal 35
3541. Is this right? I plug in the numbers like it says it I still don't get the right answer.
 
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  • #3
Gravitational potential energy is = GMm/r (where M is the mass of the earth)
Kinetic energy = 1/2 m v^2

You start at r=radius of Earth and end up at r=infinity.
A bit of rearranging and a subtraction and it shoudl be easy
 
  • #4
I set KE=GPE i get a velocity of 465,561 for a 10kg object. This doesn't seem right.
 
  • #5
Stratosphere said:
I set KE=GPE i get a velocity of 465,561 for a 10kg object. This doesn't seem right.

Well escape velocity is a function of radius so this question, as you stated, makes no sense...
 
  • #6
Stratosphere said:
Every time I try using the formula i have i have to solve for V^2 and I get it to equal 35
3541. Is this right? I plug in the numbers like it says it I still don't get the right answer.

What formula are you using?
 
  • #7
Janus said:
What formula are you using?


V[tex]^{2}[/tex]=[tex]\frac{2GM}{R}[/tex] to find the escape velocity of the Earth and i get, 465561.808.
 
  • #8
Stratosphere said:
[itex]V^{2}=\frac{2GM}{R}[/itex] to find the escape velocity of the Earth and i get, 465561.808.

You've got "tex" tags inside your formula. I've tried to fix tags in the quoted extract; but it is still not displaying for some reason. But the forumla looks fine to me. Just make sure all your variables are in consistent units. You give a number, but no units. Are you meaning km/hr, or m/s, or something else?

The gravitational constant G is usually given in SI units. If you do everything in meters and seconds and kilograms, it will work.
 
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  • #9
Stratosphere said:
Every time I try using the formula i have i have to solve for V^2 and I get it to equal 35
3541. Is this right? I plug in the numbers like it says it I still don't get the right answer.

Okay, from this answer, I'd say that you are using kilometers instead of meters for the radius of the Earth.
 
  • #10
Why would the mass of the object matter ?

GM for the Earth is almost exactly 400,000 km^3/s^2 (worth remembering)
The radius of the Earth is around 6400km

so v^2 = 2GM/r = 2*400,000/6400 = 225km^2/s^2
v = 11.2km/s (roughly)

edit - sorry i posted this straight after you replied, but it didn't show up (got a database error message)
 

1. What is escape velocity?

Escape velocity is the minimum speed needed for an object to break free from the gravitational pull of a celestial body, such as a planet or moon.

2. How is escape velocity calculated?

The formula for escape velocity is v = √(2GM/r), where G is the gravitational constant, M is the mass of the celestial body, and r is the distance from the center of the body to the object's starting point.

3. Is escape velocity the same for all celestial bodies?

No, escape velocity varies depending on the mass and size of the celestial body. For example, the escape velocity on Earth is higher than that of the Moon due to Earth's larger mass.

4. Can escape velocity be exceeded?

Yes, escape velocity is the minimum speed needed to escape a celestial body's gravitational pull. An object can exceed escape velocity if it has additional propulsion or is affected by other forces, such as a slingshot effect from a planet's gravity.

5. Is escape velocity the same as orbital velocity?

No, orbital velocity is the speed needed for an object to maintain a stable orbit around a celestial body, whereas escape velocity is the speed needed to completely break free from a celestial body's gravitational pull.

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