Mechanical Energy and Type of Motion

The gravitational potential energy of a mass $m$ at a distance $r$ from the center of a planet of mass $M$ is $E_p = -\frac{G\times M\times m}{r}$. The figure shows the graph of $E_p$ against $r$.

1. a) Explain what convention for the zero of energy is adopted in this expression: at what position is the potential energy equal to zero?
2. b) With this convention, reason why the potential energy of a satellite bound to the Earth is always negative.

A probe of $m = 800\ \text{kg}$ is located at $r = 7\times10^{6}\ \text{m}$ from the center of the Earth. Data: $G = 6.67\times10^{-11}\ \text{N}\times\text{m}^2/\text{kg}^2$ and $M_T = 5.97\times10^{24}\ \text{kg}$. Observe the diagram:

1. a) Calculate its gravitational potential energy, $E_p = -\frac{G\times M\times m}{r}$, and express it in scientific notation.
2. b) Explain the physical meaning of the negative sign of the result.

A satellite is initially at $r_1 = 7\times10^{6}\ \text{m}$ from the center of the Earth and then moves to $r_2 = 1.4\times10^{7}\ \text{m}$.

1. a) Without carrying out numerical calculations, reason using the formula $E_p = -\frac{G\times M\times m}{r}$ in which of the two positions its potential energy is greater.
2. b) In which of the two positions is the satellite more strongly bound to the Earth? Justify your answer using the sign and the value of $E_p$.

The escape velocity from the surface of a planet of mass $M$ and radius $R$ is $v_e = \sqrt{\frac{2\times G\times M}{R}}$. Earth data: $M_T = 5.97\times10^{24}\ \text{kg}$ and $R_T = 6.37\times10^{6}\ \text{m}$; $G = 6.67\times10^{-11}\ \text{N}\times\text{m}^2/\text{kg}^2$. Observe the diagram:

1. a) Calculate the escape velocity from the Earth's surface.
2. b) Express the result in $\text{km/s}$.

Lunar data: $M_L = 7.35\times10^{22}\ \text{kg}$ and $R_L = 1.74\times10^{6}\ \text{m}$; $G = 6.67\times10^{-11}\ \text{N}\times\text{m}^2/\text{kg}^2$.

1. a) Calculate the escape velocity from the surface of the Moon using $v_e = \sqrt{\frac{2\times G\times M}{R}}$.
2. b) Knowing that on Earth it is about $11.2\ \text{km/s}$, reason why taking off from the Moon requires much less energy than taking off from the Earth.

Data for Mars: $M_M = 6.42\times10^{23}\ \text{kg}$ and $R_M = 3.39\times10^{6}\ \text{m}$; $G = 6.67\times10^{-11}\ \text{N}\times\text{m}^2/\text{kg}^2$.

1. a) Calculate the escape velocity from the surface of Mars using $v_e = \sqrt{\frac{2\times G\times M}{R}}$.
2. b) Does the escape velocity depend on the mass of the escaping body? Reason your answer from the formula, considering two spacecraft, one of $10^{3}\ \text{kg}$ and another of $10^{5}\ \text{kg}$.

A spacecraft is already in orbit at a height equal to one Earth radius, that is, at $r = 2\times R_T$ from the center of the Earth, with $R_T = 6.37\times10^{6}\ \text{m}$. 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 escape velocity from that position, $v_e = \sqrt{\frac{2\times G\times M_T}{r}}$.
2. b) Given that from the surface it is $11.2\ \text{km/s}$, calculate the ratio between both velocities and relate that factor to the distance from the center of the Earth.

Data: $G = 6.67\times10^{-11}\ \text{N}\times\text{m}^2/\text{kg}^2$; Mars: $M_M = 6.42\times10^{23}\ \text{kg}$ and $R_M = 3.39\times10^{6}\ \text{m}$; Earth's escape velocity is $11.2\ \text{km/s}$.

1. a) Calculate the escape velocity from the surface of Mars and express it in $\text{km/s}$.
2. b) How many times greater is Earth's escape velocity than Martian escape velocity?
3. c) The minimum kinetic energy for escape is $E_c = \frac{1}{2}\times m\times v_e^2$: reason, using the previous results, how many times more energy per kilogram is needed to escape from Earth than from Mars.

A satellite in a circular orbit around the Earth has a kinetic energy $E_c = 2\times10^{9}\ \text{J}$. In a circular orbit, $E_p = -2\times E_c$ and $E_m = E_c + E_p$ hold. The diagram summarises the three energies:

1. a) Calculate its potential energy and its mechanical energy.
2. b) Check that the mechanical energy is negative and explain what this means for the motion of the satellite.

The International Space Station, of mass $m = 4.2\times10^{5}\ \text{kg}$, orbits at an altitude of about $400\ \text{km}$, that is, at $r = 6.77\times10^{6}\ \text{m}$ from the center of the Earth. The velocity of a satellite in a circular orbit is $v = \sqrt{\frac{G\times M}{r}}$. Data: $G = 6.67\times10^{-11}\ \text{N}\times\text{m}^2/\text{kg}^2$ and $M_T = 5.97\times10^{24}\ \text{kg}$. Observe the diagram:

1. a) Calculate its orbital velocity.
2. b) Calculate its kinetic energy and its potential energy, $E_p = -\frac{G\times M\times m}{r}$.
3. c) Calculate its mechanical energy and check that it satisfies $E_m = -E_c$.

A satellite of $m = 1000\ \text{kg}$ moves from a circular orbit of radius $r_1 = 7\times10^{6}\ \text{m}$ to another of radius $r_2 = 1.4\times10^{7}\ \text{m}$, double the first. Data: $G = 6.67\times10^{-11}\ \text{N}\times\text{m}^2/\text{kg}^2$ and $M_T = 5.97\times10^{24}\ \text{kg}$. Look at the diagram:

1. a) Calculate its mechanical energy in each orbit, $E_m = -\frac{G\times M\times m}{2\times r}$.
2. b) Using the result, reason whether in the higher orbit the satellite is more or less bound to the Earth.
3. c) Calculate its orbital speed in each case using $v = \sqrt{\frac{G\times M}{r}}$ and verify that when moving to a higher orbit the satellite moves more slowly.

A navigation satellite describes a circular orbit of radius $r = 2.96\times10^{7}\ \text{m}$ around the Earth and its mechanical energy is $E_m = -2.69\times10^{9}\ \text{J}$. 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) Rearrange the expression $E_m = -\frac{G\times M\times m}{2\times r}$ for the mass of the satellite and calculate it.
2. b) Calculate its potential energy taking into account that $E_p = 2\times E_m$.
3. c) Calculate its kinetic energy.

A satellite of $m = 600\ \text{kg}$ describes an elliptical transfer orbit with semi-major axis $a = 2.45\times10^{7}\ \text{m}$ around the Earth. The mechanical energy of an elliptical orbit is $E_m = -\frac{G\times M\times m}{2\times a}$. Data: $G = 6.67\times10^{-11}\ \text{N}\times\text{m}^2/\text{kg}^2$ and $M_T = 5.97\times10^{24}\ \text{kg}$. The orbit is shown in the figure.

1. a) Calculate the mechanical energy of the satellite in that orbit.
2. b) Calculate the mechanical energy it would have in a circular orbit of radius equal to $a$ and compare both results.
3. c) Using the conservation of mechanical energy and the expression for $E_p$, reason at which point of the ellipse (perigee, closest to Earth, or apogee, furthest away) the speed of the satellite is maximum.

Two satellites, of masses $m_1 = 500\ \text{kg}$ and $m_2 = 2000\ \text{kg}$, share the same circular orbit of radius $r = 7.17\times10^{6}\ \text{m}$ around the Earth. 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 orbital speed, $v = \sqrt{\frac{G\times M}{r}}$, and explain why it is the same for both satellites even though their masses are different.
2. b) Calculate the kinetic energy and mechanical energy of each satellite.
3. c) If the two satellites were to dock forming a single body in the same orbit, reason what would happen to their orbital speed and to the total mechanical energy of the system.

Three space probes have mechanical energies $E_{m,A} = -3\times10^{9}\ \text{J}$, $E_{m,B} = 0\ \text{J}$ and $E_{m,C} = +2\times10^{9}\ \text{J}$ relative to the Sun. The figure shows the three possible types of trajectory around the Sun:

1. a) State which type of trajectory each follows (closed, parabolic or hyperbolic) and justify it using the sign of its mechanical energy.
2. b) Which of them will end up moving away from the Sun indefinitely, reaching infinity with zero velocity?

A probe of $m = 800\ \text{kg}$ is at a distance of $r = 10^{9}\ \text{m}$ from the center of the Earth, moving away at a speed of $v = 4\times10^{4}\ \text{m/s}$. 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 kinetic energy at that position.
2. b) Calculate its gravitational potential energy.
3. c) Determine its mechanical energy and indicate, giving reasons, whether the probe will escape Earth's gravitational attraction or eventually return.

A probe leaves the Earth's surface with speed $v_0 = 1.2\times10^{4}\ \text{m/s}$, greater than the escape velocity ($11.2\ \text{km/s}$). 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) Verify numerically that $v_0$ is greater than Earth's escape velocity.
2. b) Apply the conservation of mechanical energy to calculate the speed with which the probe will reach infinity (residual velocity).
3. c) State what type of trajectory the probe describes relative to the Earth and justify it based on the sign of its mechanical energy.

Halley's Comet, with an approximate mass of $m = 2\times10^{14}\ \text{kg}$, passes through the furthest point of its orbit (aphelion), at $r = 5.24\times10^{12}\ \text{m}$ from the Sun, with a speed of $9.1\times10^{2}\ \text{m/s}$. The interstellar object 1I/'Oumuamua, of estimated mass $m = 10^{8}\ \text{kg}$, moves very far from the Sun at $2.6\times10^{4}\ \text{m/s}$, where the solar potential energy is practically zero. Data: $M_S = 1.99\times10^{30}\ \text{kg}$; $G = 6.67\times10^{-11}\ \text{N}\times\text{m}^2/\text{kg}^2$.

1. a) Calculate the mechanical energy of Halley relative to the Sun and classify its trajectory.
2. b) Calculate the mechanical energy of 'Oumuamua and classify its trajectory.
3. c) Relate the results to observations: Halley returns to the Sun approximately every $76$ years, whereas 'Oumuamua, which visited us in 2017, will never return.

A research satellite has a potential energy $E_{p,0} = -\frac{G\times M_T\times m}{R_T}$ when on the Earth's surface. Data: $R_T = 6.37\times10^{6}\ \text{m}$. The figure shows the graph of $E_p(r)$:

1. a) Set up the equation $E_p(r) = \frac{E_{p,0}}{2}$ and calculate at what distance from the center of the Earth the satellite's potential energy is half of its value at the surface.
2. b) What height above the surface does this distance correspond to? Express the result as a multiple of the Earth's radius.

A meteorite is very far from the Earth, practically at rest, and falls towards it. 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 between infinity (where $E_c = 0$ and $E_p = 0$) and the Earth's surface.
2. b) Calculate the speed with which it reaches the surface.
3. c) Explain why the result numerically coincides with Earth's escape velocity.

A small probe is launched vertically from the Earth's surface with an initial velocity of $v_0 = 5\times10^{3}\ \text{m/s}$, without subsequent thrust (ballistic launch). 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) Apply the conservation of mechanical energy between the surface and the point of maximum height (where the velocity is zero).
2. b) Calculate the maximum distance from the center of the Earth reached by the probe and the corresponding height above the surface.
3. c) Using the equation set up, reason what happens to the maximum height when $v_0$ approaches the escape velocity, $11.2\ \text{km/s}$.
4. d) Explain why it would not be correct to solve the problem using the expression $E_p = m\times g\times h$ assuming $g$ is constant.

A small celestial body approaches the Moon with a speed, when it is still very far away, of $v_\infty = 2\times10^{3}\ \text{m/s}$. Moon data: $M_L = 7.35\times10^{22}\ \text{kg}$ and $R_L = 1.74\times10^{6}\ \text{m}$; $G = 6.67\times10^{-11}\ \text{N}\times\text{m}^2/\text{kg}^2$.

1. a) Apply the conservation of mechanical energy to calculate the speed with which it will impact the lunar surface.
2. b) Compare the result with that of a body falling from rest (lunar escape velocity, $2.37\ \text{km/s}$) and express the ratio between both speeds.
3. c) The Moon has no atmosphere: explain why on the Moon practically every approaching body ends up impacting, whereas on Earth many meteoroids do not reach the ground.

The Voyager 1 probe left the Solar System taking advantage of the orbital motion of the planets. Sun data: $M_S = 1.99\times10^{30}\ \text{kg}$; mean Earth-Sun distance: $r = 1.5\times10^{11}\ \text{m}$; $G = 6.67\times10^{-11}\ \text{N}\times\text{m}^2/\text{kg}^2$. Look at the diagram:

1. a) Calculate the escape velocity from the Sun's gravitational attraction from a distance equal to the radius of Earth's orbit.
2. b) Calculate the orbital speed of Earth around the Sun assuming its orbit is circular, $v = \sqrt{\frac{G\times M_S}{r}}$.
3. c) Determine what additional speed a probe that already accompanies Earth in its orbit needs to escape the Sun, and explain why it should be launched in the direction and sense of Earth's orbital motion.

The orbital speed of a satellite in a low circular orbit (radius $r \approx R$) is $v_{orb} = \sqrt{\frac{G\times M}{R}}$, and the escape velocity from the surface is $v_e = \sqrt{\frac{2\times G\times M}{R}}$. Look at the diagram:

1. a) Show that $v_e = \sqrt{2}\times v_{orb}$ and calculate both values for the Earth ($M_T = 5.97\times10^{24}\ \text{kg}$, $R_T = 6.37\times10^{6}\ \text{m}$, $G = 6.67\times10^{-11}\ \text{N}\times\text{m}^2/\text{kg}^2$).
2. b) A rocket with its engine fired continuously can escape from the Earth without ever reaching $11.2\ \text{km/s}$: explain why the concept of escape velocity refers solely to a ballistic launch (a single initial impulse, with no subsequent thrust) and what happens energetically when the engine continues to supply energy during the ascent.