Work and Energy Balances in the Gravitational Field

A probe moves away from the Earth in a radial direction.

1. a) Draw in your exercise book the gravitational force on the probe and its displacement, and reason what sign the work done by the gravitational field has during the motion away.
2. b) Write down the relationship between that work and the change in potential energy, $W_{campo} = -\Delta E_p$, and explain its meaning.

You can do it on paper and send a photo with your answer.

A payload of $m = 1000\ \text{kg}$ moves from the Earth's surface ($r_1 = R_T$) to a position $r_2 = 2\times R_T$ from the center of the Earth. Data: $G = 6.67\times10^{-11}\ \text{N}\times\text{m}^2/\text{kg}^2$; $M_T = 5.97\times10^{24}\ \text{kg}$ and $R_T = 6.37\times10^{6}\ \text{m}$. Look at the diagram:

1. a) Calculate the potential energy of the system in each position, $E_p = -\frac{G\times M\times m}{r}$.
2. b) Calculate the change in potential energy, $\Delta E_p = E_{p,2} - E_{p,1}$, and explain its sign.

A probe of $m = 800\ \text{kg}$ is moved very slowly (practically without changing its kinetic energy) from $r_1 = 6.77\times10^{6}\ \text{m}$ to $r_2 = 7.17\times10^{6}\ \text{m}$, measured from the center of the Earth. The minimum external work required then coincides with the change in potential energy. 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 change in potential energy between both positions.
2. b) State what minimum external work must be done and what sign it has.

A piece of space debris slowly descends from an orbit of radius $r_1 = 7.17\times10^{6}\ \text{m}$ to another of radius $r_2 = 6.87\times10^{6}\ \text{m}$.

1. a) Reason, without performing calculations, what sign the work done by the gravitational field has during the descent.
2. b) State whether its potential energy increases or decreases and justify it using the expression $E_p = -\frac{G\times M\times m}{r}$.

A satellite describes a stable circular orbit around the Earth. Look at the diagram:

1. a) Given that the gravitational force is perpendicular to the velocity at every point of the orbit, what is the work done by the gravitational field in one complete orbit?
2. b) Do the kinetic energy, potential energy, and mechanical energy of the satellite change along that orbit? Justify your answer.

The energy that must be supplied to a mass $m$ at rest on the Earth's surface to place it in a circular orbit of radius $r$ is $E = G\times M_T\times m\times\left(\frac{1}{R_T} - \frac{1}{2\times r}\right)$. A small satellite of $m = 200\ \text{kg}$ is to be placed in an orbit at an altitude of $500\ \text{km}$ ($r = 6.87\times10^{6}\ \text{m}$). Data: $G = 6.67\times10^{-11}\ \text{N}\times\text{m}^2/\text{kg}^2$; $M_T = 5.97\times10^{24}\ \text{kg}$ and $R_T = 6.37\times10^{6}\ \text{m}$.

1. a) Calculate that energy and express it in standard form.
2. b) State which two energy "components" this energy is split into.

To place a satellite into a certain orbit, its potential energy needs to be increased by $\Delta E_p = 5\times10^{9}\ \text{J}$ and, in addition, it must be given an orbital kinetic energy of $E_c = 4\times10^{9}\ \text{J}$.

1. a) Calculate the total energy that must be supplied by the launch vehicle.
2. b) Explain why putting a satellite into orbit does not consist merely of "lifting" it up to that height.

A weather satellite of $m = 1500\ \text{kg}$ is to be placed into a geostationary orbit of radius $r = 4.22\times10^{7}\ \text{m}$, starting from rest on the Earth's surface. The required energy is $E = G\times M_T\times m\times\left(\frac{1}{R_T} - \frac{1}{2\times r}\right)$. 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 that energy.
2. b) Calculate what fraction of that energy corresponds to "lifting" (change in $E_p$) and what fraction to "orbiting" (orbital kinetic energy), and comment on the result.

A satellite in an internet constellation, of mass $m = 300\ \text{kg}$, must ascend from its parking orbit, at an altitude of $500\ \text{km}$ ($r_1 = 6.87\times10^{6}\ \text{m}$), to its operational orbit, at $800\ \text{km}$ ($r_2 = 7.17\times10^{6}\ \text{m}$). The work required to move from one circular orbit to another is $W = \frac{G\times M\times m}{2}\times\left(\frac{1}{r_1} - \frac{1}{r_2}\right)$. 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 work that must be provided by the satellite's thrusters.
2. b) Reason the positive sign of the result: where does that energy come from and where does it end up?

The thrusters of a satellite of $m = 500\ \text{kg}$ raise it from a circular orbit at an altitude of $500\ \text{km}$ ($r_1 = 6.87\times10^{6}\ \text{m}$) to another at $800\ \text{km}$ ($r_2 = 7.17\times10^{6}\ \text{m}$). 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 work provided by the thrusters, $W = \frac{G\times M\times m}{2}\times\left(\frac{1}{r_1} - \frac{1}{r_2}\right)$.
2. b) Calculate the orbital speed in each orbit, $v = \sqrt{\frac{G\times M}{r}}$, and verify that it decreases.
3. c) Explain the apparent paradox: the engines supply energy to the satellite and yet it ends up moving more slowly. Support your explanation using energy balances of $E_c$, $E_p$, and $E_m$.

At the end of their operational lifespan, geostationary satellites are raised by about $300\ \text{km}$ into the so-called "graveyard orbit" so as not to interfere with active ones. Consider a satellite of $m = 2000\ \text{kg}$ moving from geostationary orbit ($r_1 = 4.22\times10^{7}\ \text{m}$) to graveyard orbit ($r_2 = 4.25\times10^{7}\ \text{m}$). 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 work required for that last push, $W = \frac{G\times M\times m}{2}\times\left(\frac{1}{r_1} - \frac{1}{r_2}\right)$.
2. b) Calculate the energy cost in its day to put it into geostationary orbit from the surface, $E = G\times M_T\times m\times\left(\frac{1}{R_T} - \frac{1}{2\times r_1}\right)$, with $R_T = 6.37\times10^{6}\ \text{m}$.
3. c) Calculate the energy it would cost to "bring it down" from geostationary orbit to an orbit at a height of $300\ \text{km}$ ($r_3 = 6.67\times10^{6}\ \text{m}$) so that it re-enters, and reason, by comparing the three results, why the graveyard orbit is chosen.

A satellite of $m = 300\ \text{kg}$ decays due to friction with the upper atmosphere from a circular orbit at an altitude of $800\ \text{km}$ ($r_1 = 7.17\times10^{6}\ \text{m}$) to another at $500\ \text{km}$ ($r_2 = 6.87\times10^{6}\ \text{m}$). 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 work done by the gravitational field during the descent, $W_{campo} = G\times M\times m\times\left(\frac{1}{r_2} - \frac{1}{r_1}\right)$.
2. b) Calculate the change in mechanical energy of the satellite between both circular orbits, $\Delta E_m = E_{m,2} - E_{m,1}$ with $E_m = -\frac{G\times M\times m}{2\times r}$.
3. c) Explain why the two results do not coincide and what has happened to the difference.

A satellite of $m = 500\ \text{kg}$ loses height due to friction with the upper atmosphere and moves from a circular orbit at an altitude of $350\ \text{km}$ ($r_1 = 6.72\times10^{6}\ \text{m}$) to another at $250\ \text{km}$ ($r_2 = 6.62\times10^{6}\ \text{m}$). 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 change in kinetic energy between both orbits, $\Delta E_c$, using $E_c = \frac{G\times M\times m}{2\times r}$.
2. b) Calculate the change in potential energy, $\Delta E_p$.
3. c) Calculate the change in mechanical energy, $\Delta E_m$, and explain what has happened to that energy.
4. d) Calculate the orbital speed in each orbit and explain the "satellite paradox": when slowed down by friction, the satellite ends up moving faster.

A probe of $m = 500\ \text{kg}$ falls from $r_1 = 4\times R_T$ to $r_2 = 2\times R_T$, measured from the center of the Earth. 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 the change in its potential energy, $\Delta E_p = E_{p,2} - E_{p,1}$.
2. b) Calculate the work done by the gravitational field, $W_{campo} = -\Delta E_p$, and interpret its sign.

A meteorite of $m = 2\times10^{3}\ \text{kg}$ falls towards the Earth from very far away, starting practically from rest. 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) Apply conservation of energy to calculate the kinetic energy with which it would reach the surface if there were no atmosphere.
2. b) Calculate the corresponding speed.
3. c) A passenger car of $1200\ \text{kg}$ at $120\ \text{km/h}$ ($33.3\ \text{m/s}$) has a kinetic energy of about $6.7\times10^{5}\ \text{J}$: calculate how many times greater the meteorite's energy is.

In 2013 a meteorite about $20\ \text{m}$ in diameter exploded over Chelyabinsk (Russia). Consider a body of $m = 10^{7}\ \text{kg}$ approaching the Earth with a speed, when it is still far away, of $v_\infty = 1.5\times10^{4}\ \text{m/s}$. 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) Apply the conservation of mechanical energy to calculate the speed with which it reaches the Earth's surface.
2. b) Calculate its kinetic energy at that moment.
3. c) One kiloton of TNT is equivalent to $4.18\times10^{12}\ \text{J}$: express the meteorite's energy in kilotons and comment on the result.
4. d) Explain why the Earth's atmosphere acts as a protective shield, but only partially against bodies of this size.

The energy required for two different missions is to be compared for a payload of $m = 1000\ \text{kg}$ starting from rest on the Earth's surface. 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 the minimum energy for the payload to escape Earth's gravitational attraction, $E_{esc} = \frac{G\times M_T\times m}{R_T}$.
2. b) Calculate the energy required to place it in a low circular orbit ($r \approx R_T$), $E_{orb} = \frac{G\times M_T\times m}{2\times R_T}$.
3. c) Reason from the results why "escaping from Earth" costs exactly twice as much as "getting into low orbit", and what practical consequence this has for interplanetary missions.

An orbiter of $m = 1000\ \text{kg}$ around Mars must move from a low circular orbit at an altitude of $200\ \text{km}$ ($r_1 = 3.59\times10^{6}\ \text{m}$) to a high orbit of radius $r_2 = 1.7\times10^{7}\ \text{m}$. Mars data: $M_M = 6.42\times10^{23}\ \text{kg}$; $G = 6.67\times10^{-11}\ \text{N}\times\text{m}^2/\text{kg}^2$.

1. a) Calculate the work required, $W = \frac{G\times M_M\times m}{2}\times\left(\frac{1}{r_1} - \frac{1}{r_2}\right)$.
2. b) The mass of Mars is about $0.11$ times that of the Earth: reason why, for the same orbital radii and orbiter mass, orbit changes are energetically much cheaper around Mars than around the Earth.

A mission carries a payload of $m = 1000\ \text{kg}$ from rest on the Earth's surface until making it escape from the Earth, in two stages: first it is placed into a circular orbit at an altitude of $800\ \text{km}$ ($r = 7.17\times10^{6}\ \text{m}$) and from there it is given the final escape impulse. 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 mission is shown in the figure.

1. a) Calculate the energy of the first stage, $E_1 = G\times M_T\times m\times\left(\frac{1}{R_T} - \frac{1}{2\times r}\right)$.
2. b) Calculate the energy of the second stage: that which must be supplied to a satellite in circular orbit so that it escapes, $E_2 = \frac{G\times M_T\times m}{2\times r}$.
3. c) Add both energies together and verify that the total coincides with the direct escape energy from the surface, $E_{esc} = \frac{G\times M_T\times m}{R_T}$.
4. d) Reason why the total does not depend on the intermediate orbit chosen, relying on the fact that the final mechanical energy (at infinity, at rest) is the same in all cases.

A crewed spacecraft of $m = 5\times10^{3}\ \text{kg}$ departs from a low Earth circular orbit ($r_1 = 6.77\times10^{6}\ \text{m}$) toward the Moon, whose mean distance to the center of the Earth is $r_2 = 3.84\times10^{8}\ \text{m}$. 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 the mechanical energy of the spacecraft in low orbit, $E_{m,1} = -\frac{G\times M_T\times m}{2\times r_1}$.
2. b) Calculate the minimum mechanical energy it would have upon reaching the lunar distance if it arrived with zero velocity, $E_{m,2} = -\frac{G\times M_T\times m}{r_2}$.
3. c) Calculate the minimum energy that the orbit exit impulse must provide, $\Delta E = E_{m,2} - E_{m,1}$, and compare it with the energy required to escape from the Earth from that same orbit.
4. d) Reason why the actual mission needs additional energy: the spacecraft cannot remain at that point with zero velocity, but must enter lunar orbit or return.

To travel to the Moon, one must overcome the "neutral point", located approximately $d = 3.46\times10^{8}\ \text{m}$ from the center of the Earth, where the attractions of the Earth and the Moon cancel out. The mean Earth–Moon distance is $r_{TL} = 3.84\times10^{8}\ \text{m}$. A $m = 5\times10^{3}\ \text{kg}$ spacecraft starts from rest on the Earth's surface. 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 $M_L = 7.35\times10^{22}\ \text{kg}$. Look at the diagram:

1. a) Calculate the change in potential energy due to the Earth when traveling from the Earth's surface to the neutral point, $\Delta E_{p,T} = G\times M_T\times m\times\left(\frac{1}{R_T} - \frac{1}{d}\right)$.
2. b) Calculate the change in potential energy due to the Moon on the same journey, taking into account that the spacecraft goes from being $r_{TL}$ from the Moon to being $r_{TL} - d$ away: $\Delta E_{p,L} = G\times M_L\times m\times\left(\frac{1}{r_{TL}} - \frac{1}{r_{TL} - d}\right)$.
3. c) Add both contributions to obtain the minimum total energy for the journey to the neutral point, and compare it with the escape energy from the Earth's surface, $\frac{G\times M_T\times m}{R_T}$. What conclusion do you draw about the energy "cost" of reaching the Moon?