Question

A satellite of mass 1000 kg is supposed to orbit the earth at a height of 2000 km above the earth's surface. Find (a) its speed in the orbit, (b) is kinetic energy, (c) the potential energy of the earth-satellite system and (d) its time period. Mass of the earth = 6 × 1024 kg.

Answer 1

Answer:

10 x 12

Explanation:

Answer 2

A satellite of mass 1000 kg is supposed to orbit the earth at a height of 2000 km above the earth's surface.

(a) The speed of the satellite in the orbit is 6.9 km/s.

(b) Kinetic energy is $2.38 \times 10^{10} J$.

(c) Earth-satellite system’s potential energy is $-4.76 \times 10^{10} J$.

(d) The time period is 2.1 hours.  

Explanation:

(a)The satellite’s speed in it’s orbit is  $v=\sqrt{\frac{G M}{r+h}}=\sqrt{\frac{g r^{2}}{r+h}}$.

$\Rightarrow v=\sqrt{\frac{9.8 \times\left(6400 \times 10^{3}\right)^{2}}{10^{6} \times(6.4+2)}}$

So, $v=6.9 \mathrm{km} / \mathrm{s}$.

(b) The satellite’s kinetic energy is $K E=\frac{1}{2} m v^{2}$.

$=\frac{1}{2} \times 1000 \times(6.9 \times 103)^{2}$

$=2.38 \times 10^{10} J$.

(c) The satellite has the potential energy:

$P . E .=\frac{G M m}{(R+h)}$

$=-4.76 \times 10^{10} J$.

(d) Satellite’s time period is  $T=2 p i \frac{(r+h)}{v}$

$=76.6 \times 10.2 s=2.1 h$.