Question

A satellite of mass 1000 kg is launched to revolve around the earth in an orbit at a height of 270 km from the earth's surface. Kinetic energy of the satellite in this orbit is ______ x $10^{10}$ J.
(Mass of earth = $6 \times 10^{24}$ kg, Radius of earth = $6.4 \times 10^6$ m, Gravitational constant = $6.67 \times 10^{-11}$ Nm$^2$ kg$^{-2}$)

Hint:

The formula for the kinetic energy of a satellite in orbit can be derived from the gravitational potential energy, where the radius includes both the radius of the Earth and the height of the satellite above the surface.

Answer 1

Correct Answer: 3

Determine the kinetic energy of a satellite with a mass of \( m = 1000\,\text{kg} \) orbiting the Earth at an altitude of \( h = 270\,\text{km} \) above the surface.

Given Data:

\[ m = 1000\,\text{kg}, \quad M = 6 \times 10^{24}\,\text{kg}, \quad R = 6.4 \times 10^6\,\text{m}, \quad G = 6.67 \times 10^{-11}\,\text{Nm}^2/\text{kg}^2 \] \[ h = 270\,\text{km} = 270 \times 10^3\,\text{m} \]

Concept Used:

The gravitational force acting on a satellite in a circular orbit is equal to the centripetal force required for that orbit:

\[ \frac{GMm}{r^2} = \frac{mv^2}{r} \] \[ \Rightarrow v^2 = \frac{GM}{r} \]

where \( r = R + h \) represents the orbital radius.

The kinetic energy (\( K \)) of the satellite is given by:

\[ K = \frac{1}{2}mv^2 = \frac{1}{2}m\frac{GM}{r} \]

Step-by-Step Solution:

Step 1: Calculate the orbital radius:

\[ r = R + h = 6.4 \times 10^6\,\text{m} + 0.27 \times 10^6\,\text{m} = 6.67 \times 10^6\,\text{m} \]

Step 2: Substitute the orbital radius into the kinetic energy formula:

\[ K = \frac{1}{2}m\frac{GM}{r} \]

Step 3: Insert the given numerical values into the formula:

\[ K = \frac{1}{2} \times 1000\,\text{kg} \times \frac{(6.67 \times 10^{-11}\,\text{Nm}^2/\text{kg}^2)(6 \times 10^{24}\,\text{kg})}{6.67 \times 10^6\,\text{m}} \]

Step 4: Simplify the expression:

\[ \frac{GM}{r} = \frac{6.67 \times 6}{6.67} \times 10^{-11 + 24 - 6} = 6 \times 10^7\,\text{m}^2/\text{s}^2 \] \[ K = \frac{1}{2} \times 1000\,\text{kg} \times 6 \times 10^7\,\text{m}^2/\text{s}^2 = 3 \times 10^{10}\,\text{J} \]

Final Computation & Result:

The kinetic energy of the satellite is:

\[ \boxed{K = 3 \times 10^{10}\,\text{J}} \]

Therefore, the kinetic energy is 3 × 10¹⁰ J.