15:I[8216,["972","static/chunks/972-b28da0de92681928.js","47","static/chunks/47-a7f06e9eda1ac756.js","115","static/chunks/app/%5Bexam%5D/questions/%5Byear%5D/%5Bsubject%5D/%5Bchapter%5D/%5BquestionId%5D/page-7b6f0a12b9cc0206.js"],"BackButton"] 16:I[1553,["972","static/chunks/972-b28da0de92681928.js","47","static/chunks/47-a7f06e9eda1ac756.js","115","static/chunks/app/%5Bexam%5D/questions/%5Byear%5D/%5Bsubject%5D/%5Bchapter%5D/%5BquestionId%5D/page-7b6f0a12b9cc0206.js"],"QuestionViewer"] 17:T78b,

Here's how to determine the kinetic energy (K) and total energy (E) of the satellite:

1. Angular Momentum and Velocity:

The angular momentum (J) of a satellite in circular orbit is given by:

$$J = I\omega = mR^2 \omega$$

where:

m is the mass of the satellite
R is the orbital radius
ω is the angular velocity

We can relate angular velocity (ω) to linear velocity (v) using:

$$v = Rω$$

2. Kinetic Energy:

The kinetic energy of the satellite is:

$$K = \frac{1}{2}mv^2$$

Substituting v = Rω and J = mR²ω, we get:

$$K = \frac{1}{2}m(Rω)^2 = \frac{1}{2}m(J^2 / m^2R^4) = \frac{J^2}{2mR^2}$$

3. Total Energy:

The total energy (E) of the satellite is the sum of its kinetic energy (K) and potential energy (U). The potential energy in a gravitational field is:

$$U = -\frac{GMm}{R}$$

where G is the gravitational constant and M is the mass of the Earth.

The total energy is then:

$$E = K + U = \frac{J^2}{2mR^2} - \frac{GMm}{R}$$

We can express the gravitational constant (G) and the mass of the Earth (M) in terms of the acceleration due to gravity (g) at the Earth's surface:

$$GM = gR^2$$

Substituting this into the equation for total energy:

$$E = \frac{J^2}{2mR^2} - \frac{gR^2m}{R} = \frac{J^2}{2mR^2} - gRm$$

Now, using the centripetal acceleration equation, g = v²/

Gravitation

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