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

We are asked to find the kinetic energy of a satellite of mass \( m = 1000\,\text{kg} \) revolving around the Earth at a height \( h = 270\,\text{km} \) 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:

For a satellite in a circular orbit, the gravitational force provides the centripetal force:

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

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

The kinetic energy of the satellite is:

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

Step-by-Step Solution:

Step 1: Compute the orbital radius:

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

Step 2: Substitute in the kinetic energy formula:

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

Step 3: Substitute numerical values:

\[ K = \frac{1}{2} \times 1000 \times \frac{(6.67 \times 10^{-11})(6 \times 10^{24})}{6.67 \times 10^6} \]

Step 4: Simplify step by step:

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

Final Computation & Result:

The kinetic energy of the satellite is:

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

Hence, the required answer is 3 × 10¹⁰ J.

Answer 2

\[ KE = \frac{1}{2} mv^2 = \frac{GMm}{2r} \] \[ KE = \frac{GMm}{2(r_e + h)} \] \[ KE = \frac{6.67 \times 10^{-11} \times 6 \times 10^{24} \times 1000 \times 6.4 \times 10^6}{2(6.4 \times 10^6 + 2.7 \times 10^5)} \] \[ KE = 3 \times 10^{10} \, \text{J} \]