Gravitational potential energy problem

[Tuugii]

The gravitational potential energy W of a self-gravitating sphere of mass M and radius R depends on the detailed distribution of mass within the sphere, but it is generally of order of magnitude -GM^2/R. For the Sun, W_sun = -2G(M_sun)^2/R. What is the timescale
t = -(1/2)(W_sun)/(L_sun) over which gravitational contraction could have supplied the power radiated by the sun at it's present rate?

please give some hints, thanks a lot.

 

[Shooting Star]

I am giving some answer because I think the question belongs to the Advanced Physics section.

There is something called a “virial theorem” which holds good for self gravitating bodies and many other systems. If the system is roughly in equilibrium, so that the time averages of kinetic and potential energies are changing slowly, the virial theorem implies that T = -(1/2)V. As the star shrinks, the energy is radiated away so that the above relation is valid. So, knowing the present luminosity of the sun, we can roughly find how long it can radiate at the present rate using this mechanism.

In fact, this was the theory proposed before nuclear reactions were known about.

 

[Moetasim]

The timescale t is the amount of time it would take for the gravitational contraction of the Sun to supply the power radiated at its present rate. To solve for t, we can use the equation for gravitational potential energy and the equation for luminosity, which is the rate at which the Sun radiates energy.

First, let's rearrange the equation for gravitational potential energy to solve for R, which is the radius of the Sun:
R = -GM^2/W_sun

Next, we can substitute this value for R into the equation for the timescale:
t = -(1/2)(W_sun)/(L_sun) = -(1/2)(-GM^2/W_sun)/(L_sun)

Simplifying this further, we get:
t = (GM^2)/(2L_sunW_sun)

Now, we can substitute in the values given for the Sun's mass and luminosity:
t = (6.67*10^-11 m^3/kg*s^2 * (1.99*10^30 kg)^2) / (2 * (3.828*10^26 W) * (-2*6.67*10^-11 m^3/kg*s^2 * (1.99*10^30 kg)^2) / (6.67*10^-11 m^3/kg*s^2 * (1.99*10^30 kg)^2))

Simplifying this further, we get:
t = (3.963*10^60 m^3/kg^2*s^2) / (-1.592*10^61 m^3/kg^2*s^2)

Finally, we can cancel out the units and solve for t:
t = -0.0248 seconds

This timescale indicates that it would take a very short amount of time for gravitational contraction to supply the power radiated by the Sun at its present rate. This is due to the large mass and radius of the Sun, which results in a very high gravitational potential energy.