Question

If \( K \) is the kinetic energy of a satellite at a height \( h \) from the surface of earth, then its total energy is

• A. \( -K \)
• B. \( 2K \)
• C. \( \sqrt{K} \)
• D. \( -\dfrac{K}{2} \)
• E. \( -\dfrac{K}{4} \)

Hint:

For a satellite in circular orbit, always remember: \[ U=-2K \quad \text{and} \quad E=K+U=-K \] These two relations make orbital energy questions very quick to solve.

Answer 1

The Correct Option is A

Step 1: Recall the energy relations for a satellite in circular orbit.
For a satellite moving in a circular orbit of radius \( r \), the gravitational force provides the necessary centripetal force.
The standard energy relations are: \[ K=\frac{GMm}{2r} \] and \[ U=-\frac{GMm}{r} \] where \( K \) is kinetic energy and \( U \) is gravitational potential energy.

Step 2: Write the expression for total mechanical energy.
The total energy \( E \) of the satellite is the sum of kinetic and potential energies: \[ E=K+U \] Substituting the standard expressions, \[ E=\frac{GMm}{2r}-\frac{GMm}{r} \]

Step 3: Simplify the total energy expression.
Take \( \dfrac{GMm}{2r} \) as common: \[ E=\frac{GMm}{2r}-\frac{2GMm}{2r} \] \[ E=-\frac{GMm}{2r} \]

Step 4: Compare total energy with kinetic energy.
We already know that \[ K=\frac{GMm}{2r} \] Hence, \[ E=-\frac{GMm}{2r}=-K \]

Step 5: Note that the height \( h \) only changes the orbital radius.
If the satellite is at height \( h \) above the earth’s surface, then \[ r=R+h \] where \( R \) is the radius of the earth.
But the relation \[ E=-K \] remains true for any circular orbit, irrespective of the value of \( h \).

Step 6: Physical interpretation.
The total energy of a bound satellite is always negative.
Its magnitude equals the kinetic energy, which means: \[ |E|=K \] So the total energy is exactly the negative of the kinetic energy.

Step 7: Final conclusion.
Therefore, the total energy of the satellite is \[ \boxed{-K} \] Hence, the correct option is \[ \boxed{(1)\ -K} \]