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If the time period of a satellite is T, its kinetic energy is proportional to KE directly proportional to T -1/2 KE directly proportional to T -2/3 KE directly proportional to T 3/2 d.None of these

If the time period of a satellite is T, its kinetic energy is proportional to
 
  1. KE directly proportional to T -1/2
  2. KE directly proportional to T -2/3
  3. KE directly proportional to T 3/2
  4. d.None of these
 

Grade:12th pass

2 Answers

Piyush Kumar Behera
417 Points
4 years ago
Answer of the question is C.
The solution of the question is the given in the image below.
I hope it helps!!
Khimraj
3007 Points
4 years ago
Let KE = x
Centripetal force = C= mv²/r 
Gravitational force:= g=GMm/r²
 => C:g= mv²/r : GMm/r²  ⇒v² = GM/r  
=> x = ½m(GM/r) = GM/2r  
Here this shows that x(kinetic energy of the satellite) is inversely proportional to the radius of the satellite orbit: 
⇒x ∝ r ⁻¹  

We have time of revolution given as: 
T = 2π/ω  
Centripetal Force= Fc
Gravitational Force= Fg
Fc = Fg  
mv²/r = GMm/r²
m(ωr)²/r = GMm/r²  
m(ω²)r = GMm/r²  
ω² = GM/r³  

Now we have,
T²= 4π²/ω² 
T²=4π²/(GM/r³) = 4π²r³/GM  
The above equation shows that the square of the time period is proportional to the cube  of the radius such that:
 T² ∝ r³ (also known as Kepler's 3rd Law) => T ∝ r^3/2  

If T = c1*r^3/2 
where we hace c1 as some constant, 
and r = c2*x⁻¹, then:  
T = c1*(c2x⁻¹)^3/2 = (c1*(c2^3/2))*x^-3/2 = c3*x^-3/2, 
thus we achieve the following relation of time period revolution:
⇒ T ∝ x^-3/2 
so KE ∝ T-2/3

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