Exercises: Work and energy balances

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 centre 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 reason its sign.

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.

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 centre 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.

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 probe of $m = 500\ \text{kg}$ falls from $r_1 = 4\times R_T$ to $r_2 = 2\times R_T$, measured from the centre 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 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?

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 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.

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.

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 $E_m$.

A launch vehicle places a payload of $m = 400\ \text{kg}$ into a circular orbit at an altitude of $600\ \text{km}$ ($r = 6.97\times10^{6}\ \text{m}$), starting from rest on the 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 the total energy.
2. b) Calculate the energy per kilogram of payload and express it in standard form.
3. c) The escape energy from the surface is $6.25\times10^{7}\ \text{J}$ per kilogram: calculate what fraction the orbital launch energy represents relative to the escape energy.

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 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.

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.

For small heights compared to the Earth's radius, the change in potential energy is usually approximated by $\Delta E_p \approx m\times g_0\times h$, with $g_0 = 9.8\ \text{m/s}^2$. Consider a payload of $m = 1000\ \text{kg}$ raised $h = 100\ \text{km} = 10^{5}\ \text{m}$ above the 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 $R_T = 6.37\times10^{6}\ \text{m}$. The figure compares both expressions:

1. a) Calculate the exact change in potential energy, $\Delta E_p = G\times M_T\times m\times\left(\frac{1}{R_T} - \frac{1}{R_T + h}\right)$.
2. b) Calculate the approximate value $m\times g_0\times h$.
3. c) Determine the relative error of the approximation as a percentage.
4. d) Repeat the comparison for $h = R_T$ (a height equal to the Earth's radius) and conclude in which range of heights the approximation $m\times g_0\times h$ is acceptable.

The energy cost of placing the same payload of $m = 500\ \text{kg}$ into a low orbit at an altitude of $100\ \text{km}$ from the surface of the Earth and from the surface of the Moon is to be compared. Data: $G = 6.67\times10^{-11}\ \text{N}\times\text{m}^2/\text{kg}^2$; Earth: $M_T = 5.97\times10^{24}\ \text{kg}$, $R_T = 6.37\times10^{6}\ \text{m}$; Moon: $M_L = 7.35\times10^{22}\ \text{kg}$, $R_L = 1.74\times10^{6}\ \text{m}$.

1. a) Calculate the energy required from the Earth, $E = G\times M_T\times m\times\left(\frac{1}{R_T} - \frac{1}{2\times r}\right)$, with $r = R_T + 10^{5}\ \text{m}$.
2. b) Calculate the energy required from the Moon, with $r = R_L + 10^{5}\ \text{m}$.
3. c) Calculate the ratio between both energies and comment on what consequence the fact that taking off from the Moon is energetically so "cheap" would have for the logistics of a future lunar base.

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.

To travel to the Moon, one must overcome the "neutral point", located approximately $d = 3.46\times10^{8}\ \text{m}$ from the centre 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 travelling 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?

Unit challenge. How can we connect the fall of an object to satellite launches? Become a space mission engineer from home:

1. Home measurement and uncertainty: Drop a small ball from 5 different heights between $0.50\text{ m}$ and $2.00\text{ m}$. Record the fall time with your mobile phone (filming in slow motion and analysing the video frame by frame) (3 repetitions per height), associating its uncertainty ($\pm \Delta h$, $\pm \Delta t$). Prepare a table with the mean values and plot $v^2$ versus $h$ to determine the local value of $g$ using a mechanical energy balance ($mgh = \frac{1}{2}mv^2$).

2. Orbital energy balance: Look up the Earth's mean radius ($R_T = 6371\text{ km}$) on the Ordnance Survey Ireland (OSi) website and calculate the mass of the Earth $M_T$ from your $g$. Determine the work required to transfer a $3.5\text{ kg}$ nanosatellite from the surface to a circular orbit at an altitude of $400\text{ km}$.

3. Reasoned proposal and trajectories: Write a short report including a table, a graph, and a theoretical discussion: classify the type of motion (elliptical orbit, parabola, or hyperbola) according to the sign of the total mechanical energy ($E_m < 0$, $E_m = 0$, $E_m > 0$) and calculate the escape velocity.

Grading criteria:

• Rigor in experimental data processing, estimation of uncertainties, and graphical fit.

• Correctness in mechanical energy balances and calculation of gravitational work.

• Physical coherence when linking the sign of $E_m$ to the trajectory and gravitational escape.