Escape velocity & Black-Body Radiation

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Homework Help Overview

The discussion revolves around the concepts of escape velocity and black-body radiation. Participants are exploring the derivation of escape velocity, questioning the dependence on mass, and discussing the implications of energy conservation equations. Additionally, there is inquiry into the characteristics of the black-body radiation curve, particularly regarding the peak intensity and its origin.

Discussion Character

  • Conceptual clarification, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants are examining the equations for escape velocity, specifically whether it is derived from kinetic and potential energy equations and how mass factors into these equations. There is also a discussion about the implications of using complex numbers in calculations related to escape velocity. Questions are raised about the nature of the black-body radiation curve and the significance of its peak intensity.

Discussion Status

The discussion is active, with participants providing insights and questioning assumptions about the derivation of escape velocity. Some participants suggest that mass does not affect the escape velocity, while others are exploring the implications of energy conservation. The inquiry into black-body radiation remains open, with participants seeking to understand the reasons behind the observed intensity peak.

Contextual Notes

There are references to high school physics curriculum constraints, including the level of mathematics involved and the simplifications present in educational materials. Participants express confusion stemming from differing resources, which may affect their understanding of the concepts discussed.

f3nr15
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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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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 ?
 
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
 
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).
 
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
 
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.
 

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