Respuesta :
Answer:
Change in Kinetic Energy is [tex]- 3.44\times 10^{11}\ J[/tex]
Solution:
As per the question:
Mass of space craft, [tex]m = 1.00\times 10^{3}\ kg[/tex]
Mass of the sun, M = [tex]1.99\times 10^{30}\ kg[/tex]
Radius of the orbit, R = [tex]1.50\times 10^{11}\ m[/tex]
Radius of smaller orbit around the sun, r = [tex]1.08\times 10^{11}\ m[/tex]
Now,
We know that the Gravitational Potential Energy is given by:
[tex]U = - \frac{Gm_{1}m_{2}}{r}[/tex]
where
G = Gravitational constant
r = distance of the space craft from the center of the sun
Now,
Initial Gravitational Potential Energy is:
[tex]U_{in} = - \frac{GMm}{R}[/tex]
[tex]U_{in} = - \frac{6.67\times 10^{- 11}\times 1.99\times 10^{30}\times 1.00\times 10^{3}}{(1.50\times 10^{11})}[/tex]
[tex]U_{in} = - 8.849\times 10^{11}\ J[/tex]
Now,
Final Gravitational Potential Energy is:
[tex]U_{f} = - \frac{GMm}{r}[/tex]
[tex]U_{in} = - \frac{6.67\times 10^{- 11}\times 1.99\times 10^{30}\times 1.00\times 10^{3}}{(1.08\times 10^{11})}[/tex]
[tex]U_{f} = - 1.229\times 10^{12}\ J[/tex]
Change in the Gravitational Potential Energy equals the change in the kinetic energy of the space craft.
[tex]\Delta U = U_{f} - U_{i}[/tex]
[tex]\Delta U = - 1.229\times 10^{12} - (- 8.849\times 10^{11}) = - 3.44\times 10^{11}\ J[/tex]
Thus
[tex]\Delta KE = - 3.44\times 10^{11}\ J[/tex]
