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 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.

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 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 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.

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.

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 $E_m$.

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.

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 $m = 300\ \text{kg}$ 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 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.