(a) Many communication satellites are placed in a circular orbit around the Earth at a radius where the period (the time to go around the Earth once) is\(24\;{\rm{h}}\). If the satellite is above some point on the equator, it stays above that point as the Earth rotates, so that as viewed from the rotating Earth the satellite appears to be motionless. That is why you see dish antennas pointing at a fixed point in space. Calculate the radius of the orbit of such a "synchronous" satellite. Explain your calculation in detail.

(b) Electromagnetic radiation including light and radio waves travels at a speed of\(3 \times {10^8}\;{\rm{m}}/{\rm{s}}\). If a phone call is routed through a synchronous satellite to someone not very far from you on the ground, what is the minimum delay between saying something and getting a response? Explain. Include in your explanation a diagram of the situation.

(c) Some human-made satellites are placed in "near-Earth" orbit, just high enough to be above almost all of the atmosphere. Calculate how long it takes for such a satellite to go around the Earth once, and explain any approximations you make.

(d) Calculate the orbital speed for a near-Earth orbit, which must be provided by the launch rocket. (The advantages of near-Earth communications satellites include making the signal delay unnoticeable, but with the disadvantage of having to track the satellites actively and having to use many satellites to ensure that at least one is always visible over a particular region.)

(e) When the first two astronauts landed on the Moon, a third astronaut remained in an orbiter in circular orbit near the Moon's surface. During half of every complete orbit, the orbiter was behind the Moon and out of radio contact with the Earth. On each orbit, how long was the time when radio contact was lost?

The period of the satellite's orbit around the Earth is \(T = 24\;{\rm{h}}\).

The gravitational constant is \(G = 6.67 \times {10^{ - 11}}\;{\rm{N}} \cdot {{\rm{m}}^2}/{\rm{k}}{{\rm{g}}^2}\).

The net force on an object is equal to the rate of change of momentum and can be written as the sum of two parts. The parallel rate of change of momentum and the perpendicular rate of change of momentum\({(d\vec p/dt)_ \bot }\)are the two parts that we are concerned with.

Change in momentum is given by,

\(\frac{{d\vec p}}{{dt}} = {\frac{{d\vec p}}{{dt}}_{||}} + {\frac{{d\vec p}}{{dt}}_ \bot }{\rm{ }}\) ……………………… (1)

The object's speed is affected by the parallel rate of change of momentum, and because the speed is constant, the parallel rate is zero and equal to the rate of change of the magnitude of the momentum.

\({\frac{{d\vec p}}{{dt}}_{||}} = \frac{{d|\vec p|}}{{dt}}\hat p = 0\)

The direction shift generated by the perpendicular rate of change is known as the rate change. The quantity of the perpendicular rate change matches the rate change of the direction of the momentum at speeds significantly slower than the speed of light. .

\({\frac{{d\vec p}}{{dt}}_ \bot } = |\vec p|\frac{{d\hat p}}{{dt}} = \frac{{m{v^2}}}{R}{\rm{ }}\) ………………………. (2)

It will also be equivalent to the rate of change of momentum.

Where\(m\)is the mass of the satellite. Also, there is a gravitational force between the Earth and the satellite which is perpendicular to the Earth and causes the change in the direction of the momentum of the satellite and it is given by

\(F = G\frac{{Mm}}{{{R^2}}}\) …….…….. (3)

Where\(G\)is the gravitational constant and equals \(6.67 \times {10^{ - 11}}\;{\rm{N}} \cdot {{\rm{m}}^2}/{\rm{k}}{{\rm{g}}^2}\),\(M\)is the mass of the Earth and\(m\)is the mass of the satellite.

Both forces are the same, the next can be determined from equations (2) and (3)

\(\begin{aligned}{}\frac{{m{v^2}}}{R} = G\frac{{Mm}}{{{R^2}}}\\{v^2} = \frac{{GM}}{R}\\v = \sqrt {\frac{{GM}}{R}} \end{aligned}\) ………………… (4)

The satellite moves in almost a circle path and its speed could be calculated due to the change in the distance with time.

So, it is given by

\(v = \frac{d}{T}\) ………………… (5)

Where \(T\) is the time taken to complete one round. When the satellite travels through one round, it moves above the circumference of a circle.

(a)

The circumference is given by,

\(2\pi R\)

Where\(R\)is the radius of the kissing circle or the distance between the Earth and the satellite. So, the distance where the satellite travels is

\(d = 2\pi R\)

Plug expression for\(d\)into equation (5) to get the new form

\(v = \frac{{2\pi R}}{T}\) ……………………… (6)
From equations (4) and (6) we could get the time \(T\) as next

\(\begin{aligned}{}{v_{earth}} = {v_{satellite}}\\\sqrt {\frac{{GM}}{R}} = \frac{{2\pi R}}{T}\\\frac{{GM}}{R} = {\left( {\frac{{2\pi R}}{T}} \right)^2}\end{aligned}\)

\(\begin{aligned}{}\\R = \sqrt(3){{\frac{{GM{T^2}}}{{4{\pi ^2}}}}}\end{aligned}\) ………………….. (7)

Now plug the values for\(G\),\(M\)and\(T\)into equation (7) to get\(R\)

\(\begin{aligned}{}{\rm{R = }}\sqrt({\rm{3}}){{\frac{{{\rm{GM}}{{\rm{T}}^{\rm{2}}}}}{{{\rm{4}}{{\rm{\pi }}^{\rm{2}}}}}}}\\{\rm{ = }}\sqrt({\rm{3}}){{\frac{{\left( {{\rm{6}}{\rm{.67 \times 1}}{{\rm{0}}^{{\rm{ - 11}}}}{\rm{\;N \times }}{{\rm{m}}^{\rm{2}}}{\rm{/k}}{{\rm{g}}^{\rm{2}}}} \right)\left( {{\rm{6 \times 1}}{{\rm{0}}^{{\rm{24}}}}{\rm{\;kg}}} \right){{{\rm{(24\;h \times 3600\;s/h)}}}^{\rm{2}}}}}{{{\rm{4}}{{\rm{\pi }}^{\rm{2}}}}}}}\\{\rm{ = 42 \times 1}}{{\rm{0}}^{\rm{6}}}{\rm{\;m}}\end{aligned}\)

Thus, the radius of the orbit is \({\rm{42 \times 1}}{{\rm{0}}^{\rm{6}}}{\rm{\;m}}\).

(b)

The electromagnetic radiation travels a distance\(2d\)as the distance between the surface of the Earth and the satellite is\(d\)as shown in the figure below.

As shown by figure b below, the distance\(d\)is given by

\(d = R - {R_E}\)

Where\(R\)is the distance that we calculated in part (a) and\({R_E}\)is the radius of the Earth and equals\(6.4 \times {10^6}\;{\rm{m}}\).

The speed could be calculated due to the change in the distance with time.

So, it is given by

\(\begin{aligned}{c}v = \frac{{2d}}{T}T\\ = \frac{{2d}}{v}T\\ = \frac{{2\left( {R - {R_E}} \right)}}{v}\end{aligned}\)

Now plug our values for\(R,{R_E}\)and\(v\)into equation (1) to get\(T\)where\(v\)is the speed of the electromagnetic radiation and equals\(3 \times {10^8}\;{\rm{m}}/{\rm{s}}\)

\(\begin{aligned}{}T &= \frac{{2\left( {R - {R_E}} \right)}}{v}\\ &= \frac{{2\left( {42 \times {{10}^6}\;m - 6.4 \times {{10}^6}\;m} \right)}}{{3 \times {{10}^8}\;m/s}}\\ &= 0.24\;s\end{aligned}\)

Therefore, the minimum delay between saying something and getting a responseis \(0.24\;s\).

(C)

Near-Earth orbit means the distance between the satellite and the Earth equals the radius of the Earth\({R_E} = 6.4 \times {10^6}\;{\rm{m}}\).

So, use equation (7) in part (a) to get the time\(T\)but use\({R_E}\)instead of\(R\)

\(\begin{aligned}{}{R_E} &= \sqrt(3){{\frac{{GM{T^2}}}{{4{\pi ^2}}}}}\\R_E^3 &= \frac{{GM{T^2}}}{{4{\pi ^2}}}4{\pi ^2}\\R_E^3 &= GM{T^2}\\T &= \sqrt {\frac{{4{\pi ^2}R_E^3}}{{GM}}} \end{aligned}\)

Now plug values for\({R_E},G\)and\(M\)into equation (1) to get\(T\)

\(\begin{aligned}{}T{\rm{ }} &= \sqrt {\frac{{4{\pi ^2}R_E^3}}{{GM}}} \\ &= \sqrt {\frac{{4{\pi ^2}{{\left( {6.4 \times {{10}^6}\;m} \right)}^3}}}{{\left( {6.67 \times {{10}^{ - 11}}\;N \times {m^2}/k{g^2}} \right)\left( {6 \times {{10}^{24}}\;kg} \right)}}} \\ &= 5085\;s\\ &= 85\;\min \end{aligned}\)

So, time taken for such a satellite to go around the Earth onceis \(85\;\min \).

(d)

The satellite moves in almost a circular path over the circumference of the Earth and its speed could be calculated due to the change in the distance with time.

So, it is given by

\(v = \frac{d}{T}\)

Where\(T\)is the time taken to complete one round. When the satellite travels through one round, it moves above the circumference of a circle. The circumference is given by

\(2\pi {R_E}\)

Where\({R_E}\)is the radius of the Earth. So, the distance where the satellite travels is

\(d = 2\pi {R_E}\)

Plug expression for\(d\)into equation (1) to get the new form

\(v = \frac{{2\pi {R_E}}}{T}\)

Now plug values for\({R_E}\)and\(T\)into equation (2) to get\(v\)near-Earth

\(\begin{aligned}{}v &= \frac{{2\pi {R_E}}}{T}\\ &= \frac{{2\pi \left( {6 \times {{10}^6}\;m} \right)}}{{(5085\;s)}}\\ &= 7413\;m/s\end{aligned}\)

Thus, the orbital speed for a near-Earth orbit is\(7413\;m/s\).

(e)

The signal is lost at the distance equals the radius of the moon. So, use the expression of\(T\)as shown in part (c) to get the time but plug the values for the radius of the moon\({R_m}{ = ^\prime }1.74 \times {10^6}\;{\rm{m}}\)and the mass of the moon\({M_m} = 7.35 \times {10^{22}}\;{\rm{kg}}\)

\(\begin{aligned}{}T &= \sqrt {\frac{{4{\pi ^2}R_m^3}}{{G{M_m}}}} \\ &= \sqrt {\frac{{4{\pi ^2}{{\left( {1.74 \times {{10}^6}\;m} \right)}^3}}}{{\left( {6.67 \times {{10}^{ - 11}}\;N \times {m^2}/k{g^2}} \right)\left( {7.35 \times {{10}^{22}}\;kg} \right)}}} \\ &= 6513\;s\\ &= 109\;\min \end{aligned}\)

Hence, the duration of time for whichthe radio contact was lostis\(109\;{\rm{min}}\).