Satellites and Celestial Bodies

The International Space Station (ISS) orbits at a height of $400\ \text{km}$ above the Earth's surface. The orbital speed of a satellite in a circular orbit is $v = \sqrt{\frac{G\times M_T}{r}}$. Data: $G = 6.67\times10^{-11}\ \text{N}\times\text{m}^2/\text{kg}^2$, mass of the Earth $M_T = 5.97\times10^{24}\ \text{kg}$ and Earth radius $R_T = 6.37\times10^{6}\ \text{m}$. The diagram shows the station's orbit (not to scale):

1. a) Express the radius of the orbit, $r = R_T + h$, in meters.
2. b) Calculate the orbital speed of the ISS, in $\text{m/s}$, and express it in $\text{km/s}$.

The Moon, Earth's natural satellite, orbits at a mean distance between centers of $r = 3.84\times10^{8}\ \text{m}$. Data: $G = 6.67\times10^{-11}\ \text{N}\times\text{m}^2/\text{kg}^2$ and $M_T = 5.97\times10^{24}\ \text{kg}$.

1. a) Calculate its orbital speed using $v = \sqrt{\frac{G\times M_T}{r}}$, in $\text{m/s}$.
2. b) Compare it with the orbital speed of the ISS, about $7.7\ \text{km/s}$: which of the two bodies orbits faster, and why?

The circular orbital speed of a satellite is $v = \sqrt{\frac{G\times M}{r}}$.

1. a) Reason, from the expression, whether a satellite farther from the center of the Earth orbits faster or slower than a close one.
2. b) If the radius of the orbit is multiplied by $4$, what happens to the orbital speed? Justify your answer using the expression.

An earth-observation satellite describes a circular orbit of radius $r = 6.77\times10^{6}\ \text{m}$ with an orbital speed of $7.67\ \text{km/s}$.

1. a) Write the expression for the period in terms of radius and speed, $T = \frac{2\times\pi\times r}{v}$.
2. b) Calculate the period in seconds and express it in minutes.

The Hubble Space Telescope describes a circular orbit at a height of $540\ \text{km}$. Data: $G = 6.67\times10^{-11}\ \text{N}\times\text{m}^2/\text{kg}^2$, $M_T = 5.97\times10^{24}\ \text{kg}$ and $R_T = 6.37\times10^{6}\ \text{m}$.

1. a) Calculate its orbital speed, $v = \sqrt{\frac{G\times M_T}{R_T + h}}$.
2. b) Calculate its orbital period, $T = \frac{2\times\pi\times r}{v}$, and express it in minutes.

The International Space Station completes one orbit in $T = 92\ \text{min}$.

1. a) Calculate how many complete orbits it makes around the Earth in one day ($24\ \text{h}$).
2. b) Reason how many sunrises and sunsets the astronauts witness each day.
3. c) If the station orbited at a greater height, would it complete more or fewer orbits per day? Reason your answer using Kepler's third law.

Satellite TV dish antennas are installed fixed, always pointing at the same point in the sky.

1. a) Explain, using the characteristics of a geostationary orbit, why it is not necessary for the antenna to track the satellite.
2. b) Give two applications of geostationary satellites and reason what advantage is provided in these applications by the satellite appearing stationary.

A geostationary satellite must have a period $T = 24\ \text{h}$. Combining $v = \sqrt{\frac{G\times M_T}{r}}$ with $T = \frac{2\times\pi\times r}{v}$ gives $r^3 = \frac{G\times M_T\times T^2}{4\times\pi^2}$. Data: $G = 6.67\times10^{-11}\ \text{N}\times\text{m}^2/\text{kg}^2$, $M_T = 5.97\times10^{24}\ \text{kg}$ and $R_T = 6.37\times10^{6}\ \text{m}$. Observe the diagram (not to scale):

1. a) Express the period in seconds.
2. b) Calculate $r^3$ and, by taking the cube root, the radius of the geostationary orbit.
3. c) Calculate the height above the surface, $h = r - R_T$, and express it in $\text{km}$.

A telecommunications operator is going to place a geostationary satellite above the equator. Data: $G = 6.67\times10^{-11}\ \text{N}\times\text{m}^2/\text{kg}^2$, $M_T = 5.97\times10^{24}\ \text{kg}$, $R_T = 6.37\times10^{6}\ \text{m}$ and Earth's rotation period $T = 24\ \text{h}$.

1. a) Starting from $\frac{G\times M_T\times m}{r^2} = \frac{m\times v^2}{r}$ and $v = \frac{2\times\pi\times r}{T}$, derive the expression $r^3 = \frac{G\times M_T\times T^2}{4\times\pi^2}$.
2. b) Calculate the radius of the orbit and the altitude above the surface.
3. c) Calculate the speed of the satellite in its orbit and explain why, moving at that speed, it appears stationary when viewed from the ground.

At the Earth's surface $g_0 = 9.8\ \text{m/s}^2$, but gravity decreases with height according to $g = \frac{G\times M_T}{r^2}$. The ISS orbits with $r = 6.77\times10^{6}\ \text{m}$. Data: $G = 6.67\times10^{-11}\ \text{N}\times\text{m}^2/\text{kg}^2$ and $M_T = 5.97\times10^{24}\ \text{kg}$.

1. a) Calculate the value of $g$ in the orbit of the ISS.
2. b) Express it as a percentage of the value at the surface.
3. c) Using this result, explain why it is incorrect to state that there is "no gravity" on the ISS and what it means that the station and its crew are in continuous free fall.

Phobos, the largest of Mars's moons, orbits at only $r = 9.38\times10^{6}\ \text{m}$ from the center of the planet. Data: $G = 6.67\times10^{-11}\ \text{N}\times\text{m}^2/\text{kg}^2$ and mass of Mars $M_M = 6.42\times10^{23}\ \text{kg}$.

1. a) Calculate its orbital speed, $v = \sqrt{\frac{G\times M_M}{r}}$.
2. b) Calculate its period, $T = \frac{2\times\pi\times r}{v}$, and express it in hours.
3. c) Mars rotates on its axis in $24.6\ \text{h}$, in the same direction that Phobos orbits. Reason whether Phobos, seen from the Martian surface, rises in the east or in the west.

A student writes in an exam: "a geostationary satellite is at rest, because it does not move relative to the Earth's ground; therefore, it does not need an orbital speed nor does it have acceleration".

1. a) Identify the two errors in the reasoning.
2. b) Calculate the actual orbital speed of a geostationary satellite ($r = 4.22\times10^{7}\ \text{m}$, $T = 24\ \text{h}$), in $\text{km/s}$.
3. c) Calculate its centripetal acceleration, $a_c = \frac{v^2}{r}$, and indicate which force provides it.

In a grazing orbit (very low), the orbital radius is approximately the Earth's radius, $r \approx R_T = 6.37\times10^{6}\ \text{m}$. Data: $G = 6.67\times10^{-11}\ \text{N}\times\text{m}^2/\text{kg}^2$, $M_T = 5.97\times10^{24}\ \text{kg}$ and $g = 9.8\ \text{m/s}^2$.

1. a) Calculate the grazing orbital speed using $v = \sqrt{\frac{G\times M_T}{R_T}}$, in $\text{km/s}$.
2. b) Check that practically the same value is obtained using $v = \sqrt{g\times R_T}$ and explain, from $g = \frac{G\times M_T}{R_T^2}$, why both expressions coincide.

The Moon orbits the Earth at a mean distance $r = 3.84\times10^{8}\ \text{m}$ with an orbital speed of $1.02\ \text{km/s}$.

1. a) Calculate its orbital period using $T = \frac{2\times\pi\times r}{v}$, in seconds.
2. b) Express it in days and compare it with the duration of a lunar cycle (about four weeks).

The satellites of the European Galileo navigation system have an orbital period of $14\ \text{h}$. Use $r^3 = \frac{G\times M_T\times T^2}{4\times\pi^2}$ with $G = 6.67\times10^{-11}\ \text{N}\times\text{m}^2/\text{kg}^2$, $M_T = 5.97\times10^{24}\ \text{kg}$ and $R_T = 6.37\times10^{6}\ \text{m}$.

1. a) Express the period in seconds and calculate the radius of the orbit.
2. b) Calculate the height of the satellites above the Earth's surface.
3. c) Calculate their orbital speed, $v = \sqrt{\frac{G\times M_T}{r}}$.

Compare the orbital speed of the ISS ($r_{ISS} = 6.77\times10^{6}\ \text{m}$) with that of the Moon ($r_L = 3.84\times10^{8}\ \text{m}$) around the Earth.

1. a) Show, using $v = \sqrt{\frac{G\times M_T}{r}}$, that the ratio of speeds is $\frac{v_{ISS}}{v_L} = \sqrt{\frac{r_L}{r_{ISS}}}$.
2. b) Calculate this ratio using the data given in the question.
3. c) Explain, using Newton's law of universal gravitation, why bodies further away orbit more slowly.

An artificial satellite describes a circular orbit of radius $r$ around the Earth, of mass $M_T$.

1. a) Justify why the only force acting on the satellite is the gravitational force and why it plays the role of centripetal force; write down equation $\frac{G\times M_T\times m}{r^2} = \frac{m\times v^2}{r}$.
2. b) Deduce the orbital speed, $v = \sqrt{\frac{G\times M_T}{r}}$, and the period, $T = 2\times\pi\times\sqrt{\frac{r^3}{G\times M_T}}$, from it. Comment on why the mass $m$ of the satellite does not appear in the results.
3. c) Apply the expressions to a weather satellite in polar orbit at an altitude of $850\ \text{km}$ and calculate its speed and period. Data: $G = 6.67\times10^{-11}\ \text{N}\times\text{m}^2/\text{kg}^2$, $M_T = 5.97\times10^{24}\ \text{kg}$ and $R_T = 6.37\times10^{6}\ \text{m}$.

In the Apollo missions, the command module remained in a circular lunar orbit at an altitude of $110\ \text{km}$ while the lunar module descended to the surface. Data: $G = 6.67\times10^{-11}\ \text{N}\times\text{m}^2/\text{kg}^2$, mass of the Moon $M_L = 7.35\times10^{22}\ \text{kg}$ and lunar radius $R_L = 1.74\times10^{6}\ \text{m}$. Look at the diagram (not to scale):

1. a) Calculate the radius of the command module's orbit.
2. b) Calculate its orbital speed, $v = \sqrt{\frac{G\times M_L}{r}}$, and express it in $\text{km/s}$.
3. c) Calculate its period, in hours, and estimate how many complete orbits around the Moon the astronaut remaining on board completed each day.

For future crewed missions to Mars, placing "areostationary" communications satellites — that is, stationary relative to the Martian surface — is being studied. Data: Mars rotates in $T = 24.6\ \text{h}$; mass of Mars $M_M = 6.42\times10^{23}\ \text{kg}$; radius of Mars $R_M = 3.39\times10^{6}\ \text{m}$; $G = 6.67\times10^{-11}\ \text{N}\times\text{m}^2/\text{kg}^2$.

1. a) Reason what orbital period these satellites must have.
2. b) Calculate the radius of the areostationary orbit using $r^3 = \frac{G\times M_M\times T^2}{4\times\pi^2}$.
3. c) Calculate its altitude above the surface of Mars and compare it with the $3.6\times10^{4}\ \text{km}$ of the Earth's geostationary orbit, explaining the cause of the difference.

Suppose Earth had half its current mass, but retained its radius and its rotation period of $24\ \text{h}$. Data: $M_T = 5.97\times10^{24}\ \text{kg}$, $G = 6.67\times10^{-11}\ \text{N}\times\text{m}^2/\text{kg}^2$; the current geostationary radius is $r = 4.22\times10^{7}\ \text{m}$.

1. a) Reason, starting from $r^3 = \frac{G\times M_T\times T^2}{4\times\pi^2}$, whether the radius of the geostationary orbit would be greater than, less than, or equal to the current one.
2. b) Calculate the new geostationary radius taking into account that, since $r^3$ is proportional to $M_T$, the new radius is $r' = r\times(1/2)^{1/3}$.
3. c) Would the period still be $24\ \text{h}$? Would the satellite's orbital speed change? Justify both answers.

Kepler's third law also applies to artificial satellites. Data: ISS ($r = 6.77\times10^{6}\ \text{m}$, $T = 92\ \text{min}$) and Moon ($r = 3.84\times10^{8}\ \text{m}$, $T = 27.3$ days). The diagram compares both orbits (not to scale):

1. a) Express both periods in seconds and calculate the ratio $\frac{T^2}{r^3}$ for the ISS and for the Moon, in SI units.
2. b) Check that both ratios match (except for rounding) and explain why this must be the case.
3. c) Use this common value to predict the period of a satellite orbiting with $r = 2.0\times10^{7}\ \text{m}$ and express it in hours.

Satellite internet constellations, such as Starlink, use thousands of satellites at an altitude of $550\ \text{km}$, whereas three geostationary satellites are enough to provide coverage to almost the entire planet. Data: $G = 6.67\times10^{-11}\ \text{N}\times\text{m}^2/\text{kg}^2$, $M_T = 5.97\times10^{24}\ \text{kg}$ and $R_T = 6.37\times10^{6}\ \text{m}$. The diagram compares the two orbits (not to scale):

1. a) Calculate the speed and orbital period of a Starlink satellite.
2. b) Calculate the ratio between its period and that of a geostationary satellite ($24\ \text{h}$) and reason, using Kepler's third law, why low-orbit satellites "pass" so quickly over an observer.
3. c) Explain why a low orbit requires a constellation of thousands of satellites to provide continuous coverage, whereas three are enough in geostationary orbit.