Question

For a planet moving around the sun in an elliptical orbit, which of the following quantities remain constant?
This question has multiple correct options
$(A)$ The total energy of the sun-planet system
$(B)$ The angular momentum of the planet around the sun
$(C)$ The force of attraction between the two
$(D)$ The linear momentum of the planet.

Answer 1

Hint: According to this question we need to determine from the options which all quantities remain unchanged when a planet revolves around the sun in an elliptical orbit. By making use of the property of total energy, expression of angular momentum, gravitational force, and linear momentum, we can determine whether these quantities will remain constant or not.

Formula used:
$\dfrac{{dL}}{{dt}} = \tau $
$\tau = r \times F$
$F = G\dfrac{{{M_1}{M_2}}}{{{R^2}}}$

Complete step by step answer:

To solve this question let us individually solve the options
$(A)$ The total energy of the sun-planet system
The total energy is the sum of the energies, Kinetic, and potential energy. At different points of the planet’s rotation around the sun, we observe that the kinetic energy and potential energy individually vary across the orbit. However, the total energy that is the sum of kinetic and potential energy will remain constant throughout the rotation of the planet around the sun.
Therefore this option is correct
$(B)$ The angular momentum of the planet around the sun
Let’s consider the expression, $\dfrac{{dL}}{{dt}} = \tau $ ,
 Where L is the angular momentum and $\tau $ is the torque.
Now $\tau = r \times F$
Where r is the radius and F is the force.
This can be written as $\tau = rF\sin \theta $
The centripetal force is acting toward the center. The force and radius are at an ${\text{angle }}{0^ \circ }$.
Therefore the above expression becomes $\tau = rF\sin 0$
Now we know that $\sin ({0^ \circ }) = 0$
Therefore we get, $\tau = 0$
Substituting the value of torque in the expression $\dfrac{{dL}}{{dt}} = \tau $
We get, $\dfrac{{dL}}{{dt}} = 0$
From this expression, we can determine that the angular momentum, L, is constant for a planet revolving around the sun in an elliptical orbit.
Therefore this option is also correct.
$(C)$ The force of attraction between the planet and the sun
The gravitational force between the planet and sun is given by $F = G\dfrac{{{M_P}{M_S}}}{{{R^2}}}$
Where Mp stands for the mass of the planet, Ms is the mass of the sun, G is the gravitational constant and R is the radius (distance between planet and sun).
Since the planet is rotating in an elliptical orbit, the distance keeps changing. Therefore, the force of attraction keeps changing. Therefore this is not the correct option.
$(D)$ The linear momentum of the planet.
From the expression we have, ${\text{momentum}} = mv$
Where \[m\] is the mass and \[v\] is the velocity
Since the orbit is elliptical the velocity of the planet keeps changing from point to point. Hence, this is also not the correct option.
For a planet revolving around the sun, we have obtained that the total energy and angular momentum remain constant.

Note:
We can determine whether the angular momentum is constant also by using Kepler’s law. According to the second law, an imaginary line that connects the sun and the planet sweeps equal area in equal times. That is the areal velocity is constant. Or in other words, the angular momentum of a planet in motion around the sun in an elliptical orbit remains unchanged or constant.