Answer:
Explanation:
a. To find the gravitational potential energy you take into account the potential energy generated by a piece of mass in the disk. You also take into account that the gravitational field at point A points to the center of the disk. Then, you can consider an element of the force, Fx:
[tex]F_x=\int G\frac{mdM}{R^2+a^2}=\int G\frac{m\rho r dr d\theta }{R^2+a^2}[/tex]
m: mass of a body at a distance of "a" perpendicular to the disk.
R:radius of the disk
M: mass of the disk
G: Cavendish's constant
by solving the integral you obtain:
[tex]F_x=2\frac{GmM}{R^2}[1-cos\theta][/tex]
[tex]F_x=2GmM[1-\frac{a}{\sqrt{R^2+a^2}}][/tex] (1)
To find the gravitational energy you use:
[tex]U=-\int F_x dx=-\int[2GmM(1-\frac{x}{\sqrt{R^2+x^2}})]dx\\\\U=-2GmM[x+\sqrt{R^2+x^2}]\\\\U=-2GmM[a+\sqrt{R^2+a^2}][/tex]
you replace the values of the parameters in the point A:
[tex]U=-2(6.67*10^{-11})(250)(5.98*10^{24})[(7.83*10^6)+(\sqrt{(7.83*10^6)^2+(6.25*10^7)^2})]\\\\U=-1.41*10^{25}J[/tex]
b. For point B you have a=0.
[tex]U=-2(6.67*10^{-11})(250)(5.98*10^{24})[(7.83*10^6)+(\sqrt{((7.83*10^6)^2})]\\\\U=-3.12*10^{24}J[/tex]
c. To find the kinetic energy you use:
[tex]W_n=\Delta K\\\\F_xd=\frac{1}{2}m(v_f^2-v_o^2)[/tex]
However, the net work is also the difference between the gravitaional potential energies calculated in the previous steps.
