Question

Find the total energy and binding energy of an artificial satellite of mass 1000kg orbiting at height of 1600km above the earth's surface. Given $G=6.67\times {{10}^{-11}}N{{m}^{2}}/k{{g}^{2}}$ $R=6400km$ $M=6\times {{10}^{24}}kg$

Answer 1

At first we need to write all the values that are given in the question, now we have to find the total energy we know the formula for total energy from that derive the equation and then place the values given to find the total energy, further we also know the formula for binding energy form an equation for the binding energy and then put in the values we will get the binding energy also.

Formula used: $KE=\dfrac{GMm}{2(R+h)}$
$PE=\dfrac{-GMm}{(R+h)}$
$TE=\dfrac{GMm}{2(R+h)}+\left( \dfrac{-GMm}{(R+h)} \right)$
Binding energy of the satellite is = energy of the satellite in infinite orbit – Energy of the satellite in the given orbit.

Complete step by step answer:
In the question we are given that, the height of the satellite is 1600Km,
$G=6.67\times {{10}^{-11}}N{{m}^{2}}/k{{g}^{2}}$
$R=6400km$
$M=6\times {{10}^{24}}kg$, we are asked to find the total and the binding energy for the satellite.
We know that, total energy = KE + PE.
We know that $KE=\dfrac{GMm}{2(R+h)}$
And, $PE=\dfrac{-GMm}{(R+h)}$.
So, the total energy is $TE=\dfrac{GMm}{2(R+h)}+\left( \dfrac{-GMm}{(R+h)} \right)$, which gives us the value,
$TE=\dfrac{-GMm}{2(R+h)}$,
So, on replacing the values given in the question with the values of the formula for total energy we get,
$\Rightarrow TE=-\dfrac{6.67\times {{10}^{-11}}(6\times {{10}^{24}})(1000)}{2(6.4+1.6)\times {{10}^{6}}}$ ,
Which on solving we get,
$TE=-2.501\times {{10}^{10}}J$
Now, we know that the binding energy of the satellite is = energy of the satellite in infinite orbit – Energy of the satellite in the given orbit
$BE=0-\dfrac{-GMm}{2(R+h)}$
$BE=\dfrac{GMm}{2(R+h)}$,
So we can see that the derived formula for binding energy came the same as that of total energy except the negative sign so we can write the mod of Total energy as binding energy.
$BE=2.501\times {{10}^{10}}J$ .

Note: In the formula $KE=\dfrac{GMm}{2(R+h)}$, m is the mass of the satellite, G is the force of gravity, R is the radius of the earth and ‘h’ is the height of the satellite. While finding binding energy we solve the equation and get to know that it has just the mod value of total energy so we place the Total energy value with the opposite sign.