Question

What are:

(a) the speed and

(b) the period of a $220 kg$ satellite in an approximately circular orbit $640$ $km$ above the surface of the earth? Suppose the satellite loses mechanical energy at the average rate of $1.4 \times 10^{5} J$ per orbital revolution. Adopting the reasonable approximation that due to atmospheric resistance force, the trajectory is a circle of slowly diminishing radius. Determine the satellite’s

(c) altitude (d) speed and (e) period at the end of its $1500^{th}$ revolution. (f) Is angular momentum around the earth’s centre conserved for the satellite or the satellite-earth system.
$Universal gravitational constant, G = 6.67 \times 10^{- 11} {Nm}^{2} {kg}^{- 2}, Mass of earth, M = 5.98 \times 10^{24} kg$

Accepted Answer

(a) The magnitude of the force acting on the satellite is $\frac{G M m}{r^{2}}$ (where $km$ is the mass of earth, $m$ is the mass of the satellite and $r$ is the radius of the orbit). The force points towards the centre of the orbit is the centripetal force and here gravitational force acts as a centripetal force.

By Newton's second law,

$\frac{G M m}{r^{2}} = \frac{m {v_{o}}^{2}}{r}$ [where $v_{o}$ is the orbital speed of satellite]

And the orbital speed is given by,

$v_{o} = \sqrt{\frac{G M}{r}}$

Radius of the orbit is the sum of earth's radius and the altitude of the satellite,

$r = R + h$

$r = (6.37 \times 10^{6} + 640 \times 10^{3}) m$ [Radius of earth $R = 6.37 \times 10^{6} m$ ]

$\Rightarrow r = 7.01 \times 10^{6} m$

Thus,

$v_{o} = \sqrt{\frac{G M}{r}}$

$220 kg$ $[\begin{matrix} G = 6.67 \times 10^{- 11} {Nm}^{2} {kg}^{- 2} \\ M = 5.98 \times 10^{24} kg \end{matrix}]$

$\Rightarrow v_{o} = 7.54 \times 10^{3} m s^{- 1}$

(b) For time period,

$T = \frac{2 π r}{v_{o}}$

$\Rightarrow T = \frac{2 π \times 7.01 \times 10^{6}}{7.54 \times 10^{3}}$

$\Rightarrow T = 5.84 \times 10^{3} s \approx 97 \min$

(c) Total energy of satellite moving in an orbit of radius $r$ is

$E_{0} = \frac{- G M m}{2 r}$

The initial energy is

$E_{0} = - \frac{(6.67 \times 10^{- 11}) (5.98 \times 10^{24}) (220)}{2 (7.01 \times 10^{6})}$ [Mass of satellite $m = 220 kg$ ]

$\Rightarrow E_{0} = - 6.26 \times 10^{9} J$

Now given that the satellite loses mechanical energy at the average rate of $1.4 \times 10^{5} J$ per orbital revolution.

So the energy of the satellite in $1500^{t h}$ orbit is

$E = - 6.26 \times 10^{9} - (1500) (1.4 \times 10^{5})$

$\Rightarrow E = - 6.47 \times 10^{9} J$

We know that total energy of satellite is

$E = - \frac{G M m}{2 r}$

$∴ r = - \frac{G M m}{2 E}$

Therefore, orbital radius of $1500^{th}$ orbit is,

$r = - \frac{(6.67 \times 10^{- 11}) (5.98 \times 10^{24}) (220)}{2 (- 6.47 \times 10^{9})}$

$\Rightarrow r = 6.78 \times 10^{6} m$

Thus, altitude is

$h = r - R$

$\Rightarrow h = 6.78 \times 10^{6} m - 6.37 \times 10^{6} m$

$\Rightarrow h = 4.1 \times 10^{5} m$

(d) The speed of satellite at the end of $1500^{th}$ revolution is,

$v = \sqrt{\frac{G M}{r}}$

$\Rightarrow v = \sqrt{\frac{(6.67 \times 10^{- 11}) (5.98 \times 10^{24})}{6.78 \times 10^{6}}}$

$\Rightarrow v = 7.67 \times 10^{3} m s^{- 1} \approx 7.7 km s^{- 1}$

(e) Time period at the end of $1500^{th}$ revolution is,

$T = \frac{2 π r}{v}$

$\Rightarrow T = \frac{2 π (6.78 \times 10^{6} m)}{7.67 \times 10^{3} m s^{- 1}}$

$\Rightarrow T = 5.6 \times 10^{3} s \approx 93 \min$

(f) The law of conservation of angular momentum states that when no external torque acts on an object, no change of angular momentum will occur.

Here, in satellite motion torque acting on the satellite is due to gravitational attraction which is an internal force for satellite-earth system. Hence, angular momentum around the earth's centre is conserved.

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