Mechanical energy and type of motion

The gravitational potential energy of a mass $m$ at a distance $r$ from the centre of a planet of mass $M$ is $E_p = -\frac{G\cdot M\cdot m}{r}$. 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 $r = 7\cdot10^{6}\ \text{m}$ from the centre of the Earth. Data: $G = 6,67\cdot10^{-11}\ \text{N}\cdot\text{m}^2/\text{kg}^2$ and $M_T = 5,97\cdot10^{24}\ \text{kg}$. Observe the diagram:

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

A satellite is initially at $r_1 = 7\cdot10^{6}\ \text{m}$ from the centre of the Earth and then moves to $r_2 = 1,4\cdot10^{7}\ \text{m}$.

1. a) Without carrying out numerical calculations, reason using the formula $E_p = -\frac{G\cdot M\cdot m}{r}$ 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 $v_e = \sqrt{\frac{2\cdot G\cdot M}{R}}$. Earth data: $M_T = 5,97\cdot10^{24}\ \text{kg}$ and $R_T = 6,37\cdot10^{6}\ \text{m}$; $G = 6,67\cdot10^{-11}\ \text{N}\cdot\text{m}^2/\text{kg}^2$. 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: $M_L = 7,35\cdot10^{22}\ \text{kg}$ and $R_L = 1,74\cdot10^{6}\ \text{m}$; $G = 6,67\cdot10^{-11}\ \text{N}\cdot\text{m}^2/\text{kg}^2$.

1. a) Calculate the escape velocity from the surface of the Moon using $v_e = \sqrt{\frac{2\cdot G\cdot M}{R}}$.
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: $M_M = 6,42\cdot10^{23}\ \text{kg}$ and $R_M = 3,39\cdot10^{6}\ \text{m}$; $G = 6,67\cdot10^{-11}\ \text{N}\cdot\text{m}^2/\text{kg}^2$.

1. a) Calculate the escape velocity from the surface of Mars using $v_e = \sqrt{\frac{2\cdot G\cdot M}{R}}$.
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 satellite in a circular orbit around the Earth has a kinetic energy $E_c = 2\cdot10^{9}\ \text{J}$. In a circular orbit, $E_p = -2\cdot E_c$ 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 $m = 4,2\cdot10^{5}\ \text{kg}$, orbits at an altitude of about $400\ \text{km}$, that is, at $r = 6,77\cdot10^{6}\ \text{m}$ from the centre of the Earth. The velocity of a satellite in a circular orbit is $v = \sqrt{\frac{G\cdot M}{r}}$. Data: $G = 6,67\cdot10^{-11}\ \text{N}\cdot\text{m}^2/\text{kg}^2$ and $M_T = 5,97\cdot10^{24}\ \text{kg}$. Observe the diagram:

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

A navigation satellite describes a circular orbit of radius $r = 2,96\cdot10^{7}\ \text{m}$ around the Earth and its mechanical energy is $E_m = -2,69\cdot10^{9}\ \text{J}$. Data: $G = 6,67\cdot10^{-11}\ \text{N}\cdot\text{m}^2/\text{kg}^2$ and $M_T = 5,97\cdot10^{24}\ \text{kg}$.

1. a) Rearrange the expression $E_m = -\frac{G\cdot M\cdot m}{2\cdot r}$ for the mass of the satellite and calculate it.
2. b) Calculate its potential energy taking into account that $E_p = 2\cdot E_m$.
3. c) Calculate its kinetic energy.

Three space probes have mechanical energies $E_{m,A} = -3\cdot10^{9}\ \text{J}$, $E_{m,B} = 0\ \text{J}$ and $E_{m,C} = +2\cdot10^{9}\ \text{J}$ 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 centre of the Earth, moving away at a speed of $v = 4\cdot10^{4}\ \text{m/s}$. Data: $G = 6,67\cdot10^{-11}\ \text{N}\cdot\text{m}^2/\text{kg}^2$ and $M_T = 5,97\cdot10^{24}\ \text{kg}$.

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 $v_0 = 1,2\cdot10^{4}\ \text{m/s}$, greater than the escape velocity ($11,2\ \text{km/s}$). Data: $G = 6,67\cdot10^{-11}\ \text{N}\cdot\text{m}^2/\text{kg}^2$; $M_T = 5,97\cdot10^{24}\ \text{kg}$ and $R_T = 6,37\cdot10^{6}\ \text{m}$.

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 $m = 2\cdot10^{14}\ \text{kg}$, passes through the furthest point of its orbit (aphelion), at $r = 5,24\cdot10^{12}\ \text{m}$ from the Sun, with a speed of $9,1\cdot10^{2}\ \text{m/s}$. The interstellar object 1I/'Oumuamua, of estimated mass $m = 10^{8}\ \text{kg}$, moves very far from the Sun at $2,6\cdot10^{4}\ \text{m/s}$, where the solar potential energy is practically zero. Data: $M_S = 1,99\cdot10^{30}\ \text{kg}$; $G = 6,67\cdot10^{-11}\ \text{N}\cdot\text{m}^2/\text{kg}^2$.

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 $E_{p,0} = -\frac{G\cdot M_T\cdot m}{R_T}$ when on the Earth's surface. Data: $R_T = 6,37\cdot10^{6}\ \text{m}$. 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 centre 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: $G = 6,67\cdot10^{-11}\ \text{N}\cdot\text{m}^2/\text{kg}^2$; $M_T = 5,97\cdot10^{24}\ \text{kg}$ and $R_T = 6,37\cdot10^{6}\ \text{m}$.

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 $v_0 = 5\cdot10^{3}\ \text{m/s}$, without subsequent thrust (ballistic launch). Data: $G = 6,67\cdot10^{-11}\ \text{N}\cdot\text{m}^2/\text{kg}^2$; $M_T = 5,97\cdot10^{24}\ \text{kg}$ and $R_T = 6,37\cdot10^{6}\ \text{m}$. 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 centre 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 $E_p = m\cdot g\cdot h$ assuming $g$ is constant.

A small celestial body approaches the Moon with a speed, when it is still very far away, of $v_\infty = 2\cdot10^{3}\ \text{m/s}$. Moon data: $M_L = 7,35\cdot10^{22}\ \text{kg}$ and $R_L = 1,74\cdot10^{6}\ \text{m}$; $G = 6,67\cdot10^{-11}\ \text{N}\cdot\text{m}^2/\text{kg}^2$.

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.