Escape velocity & Black-Body Radiation

In summary: The textbook you mention does not have the correct derivation for the escape velocity. Instead, it says that the escape velocity is v = √(2GM/r). This is incorrect, and you would need to use complex numbers in order to calculate it correctly.
  • #1
f3nr15
22
0
I'm being confused between a Physics past exam paper book and a Physics study guide.

Is the escape velocity derived from EK + EP = 0 (this makes the escape velocity depend on the mass of the body to escape from i.e. Earth or a planet) or EK = EP (this makes the escape velocity depend on the mass of the escaping object) ?

(Where EK = 0.5mv2 and EP = - GmM/r)Also, the final step I must take in conquering the necessary Quantum Physics is the Black Body Radiation curve.

Why is there a very large maximum point (or peak), say at approximately 210nm then a sudden plummet in the curve after this ?
Where is this extra large intensity emitted from ?
 
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  • #2
f3nr15 said:
I'm being confused between a Physics past exam paper book and a Physics study guide.

Is the escape velocity derived from EK + EP = 0 (this makes the escape velocity depend on the mass of the body to escape from i.e. Earth or a planet) or EK = EP (this makes the escape velocity depend on the mass of the escaping object) ?

(Where EK = 0.5mv2 and EP = - GmM/r)
?? Except for a sign change, the two equations are identical. Certainly if one depends "on the mass of the body to escape from" the other must also. In either form, you can divide the entire equation by m and eliminate m.


Also, the final step I must take in conquering the necessary Quantum Physics is the Black Body Radiation curve.

Why is there a very large maximum point (or peak), say at approximately 210nm then a sudden plummet in the curve after this ?
Where is this extra large intensity emitted from ?
 
  • #3
f3nr15 said:
EK = EP
(Where EK = 0.5mv2 and EP = - GmM/r)

EK + EP = 0

That is the equation for conservation of energy. The equation you wrote above is just the same but you missed a - sign in the right hand side. Also, just use your equation and fill in the values for kinetic and potential energy. You will see that your value for the velocity is a complex number !

Also, the second conclusion on mass dependence is incorrect since m will be devided out of the two energy equations ! So, no dependence on the object's mass.

marlon
 
  • #4
marlon said:
EK + EP = 0

That is the equation for conservation of energy. The equation you wrote above is just the same but you missed a - sign in the right hand side. Also, just use your equation and fill in the values for kinetic and potential energy. You will see that your value for the velocity is a complex number !

Also, the second conclusion on mass dependence is incorrect since m will be devided out of the two energy equations ! So, no dependence on the object's mass.

marlon

High School Physics textbooks say the escape velocity is v = √(2GM/r), the highest level of Maths in High School Physics is simple trigonometry (sinθ & cosθ only), fractions (Fc=mv2/r), substitution and quadratics and definitely NO calculus (a=Δv/Δt instead e.g Final Velocity - Initial Velocity/Final Time - Initial Time)
As for Complex Numbers, that is studied in the highest level of Mathematics offered at my school (which I don't take).
 
  • #5
f3nr15 said:
High School Physics textbooks say the escape velocity is v = √(2GM/r), the highest level of Maths in High School Physics is simple trigonometry (sinθ & cosθ only), fractions (Fc=mv2/r), substitution and quadratics and definitely NO calculus (a=Δv/Δt instead e.g Final Velocity - Initial Velocity/Final Time - Initial Time)
As for Complex Numbers, that is studied in the highest level of Mathematics offered at my school (which I don't take).

If you would have done that calculation, you would have found out that you will need to take the squareroot of a negative number. This does not work unless you use complex numbers. Such numbers are unphysical in classical mechanics. That's all. It is all about the squareroot of the negative number ! Whether you know complex numbers or not is irrelevant in this case.

marlon
 
  • #6
marlon said:
If you would have done that calculation, you would have found out that you will need to take the squareroot of a negative number. This does not work unless you use complex numbers. Such numbers are unphysical in classical mechanics. That's all. It is all about the squareroot of the negative number ! Whether you know complex numbers or not is irrelevant in this case.

marlon

The derivations from both my study-guide and my teacher say:

0.5mv2 + (-GMm/r) = 0
0.5mv2 - GMm/r = 0
0.5mv2 = GMm/r
mv2 = 2GMm/r
v2 = 2GM/r
v = √(2GM/r)

Otherwise I can't calculate the escape velocity any other way.
The Physics curriculum from which we learn from has been extremely simplified.
 

1. What is escape velocity?

Escape velocity is the minimum speed that an object needs to escape the gravitational pull of a celestial body, such as a planet or moon. It is determined by the mass and radius of the body and is the speed at which the object's kinetic energy is equal to the gravitational potential energy at the surface of the body.

2. How is escape velocity calculated?

The formula for calculating escape velocity is V = √(2GM/R), where V is the escape velocity, G is the gravitational constant, M is the mass of the celestial body, and R is the radius of the celestial body. This formula assumes that there is no air resistance and the object is launched from the surface of the body.

3. What is black-body radiation?

Black-body radiation is the electromagnetic radiation emitted by a perfect black body, which absorbs all radiation that falls on it. It is thermal radiation that depends only on the temperature of the object and is emitted in a continuous spectrum. This concept is important in understanding the properties of stars and other celestial bodies.

4. How is black-body radiation related to escape velocity?

Black-body radiation plays a role in determining a celestial body's temperature, which in turn affects its escape velocity. The higher the temperature of a celestial body, the faster its particles move, and the higher its escape velocity will be. This is because the kinetic energy of the particles increases with temperature, making it more difficult for them to be held by the body's gravitational pull.

5. Can escape velocity be exceeded?

Yes, it is possible for an object to exceed escape velocity. However, this does not mean that the object will escape the celestial body's gravitational pull. Other factors, such as air resistance, can slow down the object and prevent it from escaping. Also, if the object is launched from a point above the surface of the body, it may have a lower escape velocity and be able to escape even if it does not reach the calculated velocity at the surface.

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