Question

The time period of revolution of a satellite is t then its kinetic energy is proportional to?

Answer 1

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