Question

An artificial satellite moving in a circular orbit around the earth has total energy $ E $ $ \left( {{\text{kinetic energy + potential energy}}} \right) $. Its potential energy is given by
(A) $ 2E $
(B) $ \dfrac{E}{2} $
(C) $ \dfrac{E}{4} $
(D) Data insufficient

Answer 1

Hint: The total energy of an object in a gravitational field is considered negative. The potential energy cannot be greater than the total energy i.e. kinetic plus potential energy.

Formula used: In this solution we will be using the following formula;
 $ U = - \dfrac{{GMm}}{r} $ where $ G $ is the universal gravitational constant., $ M $ is the mass of the earth, $ m $ is the mass of the satellite and $ r $ is the distance between the centre of the earth and that of the satellite.

Complete step by step answer
To find the relation of the potential energy with the total energy, we must compare their equations:
The equation of the potential energy of an object in a gravitational field is given as
 $ U = - \dfrac{{GMm}}{r} $ where $ G $ is the universal gravitational constant., $ M $ is the mass of the earth, $ m $ is the mass of the satellite and $ r $ is the distance between the centre of the earth and that of the satellite.
 $ E = - \dfrac{{GMm}}{{2r}} $ all quantities signify the same as above.
Hence, comparing both equations by dividing the potential energy by the total we have that
 $ \dfrac{U}{E} = \left( { - \dfrac{{GMm}}{r}} \right) \div \left( { - \dfrac{{GMm}}{{2r}}} \right) $
Then we get,
 $ \dfrac{U}{E} = - \dfrac{{GMm}}{r} \times - \dfrac{{2r}}{{GMm}} $
By cancelling and simplifying, we have that
 $ \dfrac{U}{E} = 2 $
 $ \Rightarrow U = 2E $

Hence, the correct answer is A.

Note
From above, the potential energy may appear like it is greater than the total energy because it is twice the total energy. However, if one considers the negative sign, it should be interpreted that the potential energy must be more negative than the total energy (since, doubling a negative number increases the negativity) hence, $ 2E < E $
Also, the potential energy comes from integrating the force with respect to the radial distance as in
 $ U = - \int_0^r F dr $
Since, $ F = - \dfrac{{GMm}}{{{r^2}}}\hat r $ where $ \hat r $ is the unit vector in the positive radial direction
Then by inserting into the integral we have
 $ U = - \int_0^r { - \dfrac{{GMm}}{{{r^2}}}\hat rdr} $
Actually, performing the integration gives us,
 $ U = - \dfrac{{GMm}}{r} $ we can drop the unit vector since potential energy is a scalar quantity.