Satellites and Binding Energy EX 1

Since the gravitational force between the Sun and Halley's comet is the only force acting in this system, and since the gravitational force is a conservative force, the mechanical energy of this system is conserved.

(a) The mechanical energy is the sum of the kinetic and potential energies. The kinetic energy of the Sun is zero and therefore the total kinetic energy of the system is equal to the kinetic energy of the comet. To determine the kinetic energy we have first to find the mass of the comet in terms of $kg$. We will assume that its shape is spherical (it is probably not true in reality but we cannot see the body of any comet--they are too small to be visible). We will convert the density to units ${\tfrac {kg}{m^{3}}}$.

$\rho =1{\tfrac {g}{cm^{3}}}=1{\frac {g{\tfrac {1kg}{1000g}}}{cm^{3}{\tfrac {(1m)^{3}}{(100cm)^{3})}}}}=1\times 10^{3}{\tfrac {kg}{m^{3}}}$

Using $m=V\times d$ we find:

$m={\frac {4}{3}}\pi r^{3}d={\frac {4}{3}}\pi (2\times 10^{4})^{3}10^{3}=3.3\times 10^{16}kg$

Therefore the kinetic energy is equal to $K={\frac {1}{2}}mv^{2}={\frac {1}{2}}(3.3\times 10^{16})(5.5\times 10^{4})^{2}=5.1\times 10^{25}J$ when it is nearest the Sun. At this distance the gravitational potential energy is equal to

$U=-{Gm_{S}m_{c} \over r}=-{(6.67\times 10^{-11})(2\times 10^{30})(3.3\times 10^{16}) \over 8.8\times 10^{10}}=-5.0\times 10^{25}J$

Therefore the total energy $E=5.1\times 10^{25}+(-5.0\times 10^{25})=1\times 10^{24}J$.

(b) We will use the law of conservation of energy again. The comet still has the energy $1\times 10^{24}J$ when it is at the farthest distance from the Sun ($5.3\times 10^{12}m$). The kinetic energy is equal to $E-U$ where

$U=-{Gm_{S}m_{c} \over r}=-{(6.67\times 10^{-11})(2\times 10^{30})(3.3\times 10^{16}) \over 5.3\times 10^{12}}=-8.3\times 10^{23}J$

We find the kinetic energy to be equal to $1.83\times 10^{24}J$. From this value we find the speed to be $1.1\times 10^{4}m/s$.