Field Strength of a System of Masses

The gravitational field $\vec{g}$ at a point is defined as the force per unit mass that would act on a test mass placed there: $\vec{g} = \frac{\vec{F}}{m}$.

1. a) From that definition, show that $1\ \text{N/kg} = 1\ \text{m/s}^2$.
2. b) Explain in your own words what it means for the field at a point in space to be equal to $g = 3,7\ \text{N/kg}$: what force would a mass of $1\ \text{kg}$ experience? What about a mass of $10\ \text{kg}$?

The gravitational field is a vector quantity.

1. a) State the direction and sense of the vector $\vec{g}$ created by the Earth at any point around it.
2. b) Draw in your notebook the gravitational field vector at eight points around the planet (a field map): what do the directions of all the vectors have in common, and how does the magnitude change when moving away from the planet?
3. c) Explain why the gravitational field is said to be central and attractive.

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

The value of gravity at the Earth's surface can be obtained from the law of universal gravitation, as shown in the figure.

1. a) Starting from the law of universal gravitation and the definition of gravitational field (force per unit mass), derive the expression for the magnitude of the gravitational field $g$ that a spherical planet of mass $M$ and radius $R$ creates at its surface.
2. b) Calculate its value for the Earth. Data: $G = 6,67\cdot10^{-11}\ \text{N}\cdot\text{m}^2/\text{kg}^2$, $M_T = 5,97\cdot10^{24}\ \text{kg}$, $R_T = 6,37\cdot10^{6}\ \text{m}$.
3. c) Compare the result with the usual value $g = 9,8\ \text{m/s}^2$.

Before sending a crewed mission to Mars, it is useful to know the gravity on its surface.

1. a) Calculate the magnitude of the gravitational field on the surface of Mars. Data: mass of Mars $M_M = 6,42\cdot10^{23}\ \text{kg}$, radius of Mars $R_M = 3,39\cdot10^{6}\ \text{m}$, $G = 6,67\cdot10^{-11}\ \text{N}\cdot\text{m}^2/\text{kg}^2$.
2. b) The radius of Mars is approximately half that of the Earth and yet its gravity is significantly less than half that of the Earth: explain why this is the case by looking at the mass.

The Apollo program astronauts found that on the Moon, objects weigh much less than on Earth.

1. a) Calculate the gravitational field on the surface of the Moon. Data: mass of the Moon $M_L = 7,35\cdot10^{22}\ \text{kg}$, lunar radius $R_L = 1,74\cdot10^{6}\ \text{m}$, $G = 6,67\cdot10^{-11}\ \text{N}\cdot\text{m}^2/\text{kg}^2$.
2. b) Calculate the ratio $\frac{g_T}{g_L}$ taking $g_T = 9,8\ \text{m/s}^2$ and check that it is approximately equal to $6$.

Two point masses, $m_1 = 5,0\ \text{kg}$ and $m_2 = 10\ \text{kg}$, are fixed on a laboratory bench at positions $x_1 = 0$ and $x_2 = 2,0\ \text{m}$, as shown in the figure.

1. a) Calculate the field created by each mass at point $x = 0,50\ \text{m}$, stating its magnitude and sense ($G = 6,67\cdot10^{-11}\ \text{N}\cdot\text{m}^2/\text{kg}^2$).
2. b) Apply the principle of superposition to obtain the total field at that point.
3. c) Reason in which direction a small test mass released there from rest would accelerate.

Three equal point masses of $6,0\ \text{kg}$ are aligned at positions $x = 0$, $x = 1,0\ \text{m}$ and $x = 3,0\ \text{m}$ on a laboratory ruler, as shown in the figure.

1. a) Calculate the field created by each of them at point $x = 2,0\ \text{m}$, stating its magnitude and sense ($G = 6,67\cdot10^{-11}\ \text{N}\cdot\text{m}^2/\text{kg}^2$).
2. b) Apply superposition and calculate the total field at that point.
3. c) Without solving the equation, reason whether there must exist a point between $x = 2,0\ \text{m}$ and $x = 3,0\ \text{m}$ where the total field cancels out: consider the sense of the total field at $x = 2,0\ \text{m}$ and what happens as you approach the mass located at $x = 3,0\ \text{m}$.

When Jupiter is closest to the Earth, it is at a distance of about $6,3\cdot10^{11}\ \text{m}$.

1. a) Calculate the gravitational field that Jupiter then creates on the Earth ($M_J = 1,90\cdot10^{27}\ \text{kg}$, $G = 6,67\cdot10^{-11}\ \text{N}\cdot\text{m}^2/\text{kg}^2$).
2. b) Calculate the force with which Jupiter attracts a $70\ \text{kg}$ person ($F = m\cdot g$) and compare it with their weight on Earth ($g = 9,8\ \text{m/s}^2$).
3. c) In light of the result, critically comment on the astrological claim that "the position of Jupiter influences people".

Between the Earth and the Moon, there is a point where their gravitational fields cancel out. The diagram shows the situation:

1. a) With the origin at the centre of the Earth, write the condition for cancellation: $G\cdot\frac{M_T}{x^2} = G\cdot\frac{M_L}{(d - x)^2}$, where $d = 3,84\cdot10^{8}\ \text{m}$ is the Earth–Moon distance.
2. b) Simplify $G$, take square roots and solve for $x$ using $M_T = 5,97\cdot10^{24}\ \text{kg}$ and $M_L = 7,35\cdot10^{22}\ \text{kg}$ (note that $\frac{M_T}{M_L} \approx 81$, whose square root is $9$).
3. c) At what distance from the Moon is that point? Reason why it is much closer to the Moon than to the Earth.

A spacecraft of $1,2\cdot10^{4}\ \text{kg}$ travels from the Earth to the Moon along the line joining their centres.

1. a) Calculate at what distance from the centre of the Earth the total gravitational field (Earth plus Moon) vanishes. Data: $M_T = 5,97\cdot10^{24}\ \text{kg}$, $M_L = 7,35\cdot10^{22}\ \text{kg}$, $d = 3,84\cdot10^{8}\ \text{m}$, $G = 6,67\cdot10^{-11}\ \text{N}\cdot\text{m}^2/\text{kg}^2$; take $\sqrt{\frac{M_T}{M_L}} \approx 9$.
2. b) What net gravitational force do the Earth and the Moon exert on the spacecraft when it is at that point?
3. c) With the engines off, the spacecraft moves $10^{6}\ \text{m}$ from that point towards the Moon: calculate the total field there (magnitude and sense) and reason whether the cancellation point is a position of stable or unstable equilibrium.

Two equal point masses, $m = 2,0\ \text{kg}$, are at points $(0; 0)$ and $(2; 0)\ \text{m}$ of the plane, as shown in the figure.

1. a) Calculate the distance of each mass to point $P(1; 1)\ \text{m}$ and the magnitude of the field created by each at $P$ ($G = 6,67\cdot10^{-11}\ \text{N}\cdot\text{m}^2/\text{kg}^2$).
2. b) Draw the two field vectors at $P$ in your notebook and reason which components cancel out by symmetry.
3. c) Knowing that each vector makes an angle of $45^\circ$ with the vertical ($\cos 45^\circ \approx 0,71$), calculate the total field at $P$: magnitude, direction and sense.

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

Three equal point masses, $m = 100\ \text{kg}$, occupy three vertices of a square of side length $L = 2,0\ \text{m}$, as shown in the figure.

1. a) Calculate the magnitude of the field created by each mass at the fourth vertex (note: the mass at the opposite vertex is at a distance $L\cdot\sqrt{2}$). Data: $G = 6,67\cdot10^{-11}\ \text{N}\cdot\text{m}^2/\text{kg}^2$.
2. b) Resolve the three field vectors along the directions of the sides of the square and calculate the components of the total field ($\cos 45^\circ \approx 0,71$).
3. c) Calculate the magnitude of the total field and reason, by symmetry, why its direction is along the diagonal.
4. d) What force would a test mass of $50\ \text{kg}$ experience if placed at that vertex?

In a modern version of the Cavendish experiment, two tungsten spheres of $500\ \text{kg}$ each are used, placed at points $(0; 0)$ and $(3; 0)\ \text{m}$.

1. a) Show that point $P(1,5; 2)\ \text{m}$ is at a distance of $2,5\ \text{m}$ from each sphere and calculate the magnitude of the field created by each sphere at $P$ ($G = 6,67\cdot10^{-11}\ \text{N}\cdot\text{m}^2/\text{kg}^2$).
2. b) The angle that each field vector makes with the perpendicular to the segment joining the spheres satisfies $\sen\theta = 0,6$ and $\cos\theta = 0,8$: reason which components cancel out and calculate the total field at $P$ (magnitude, direction and sense).
3. c) Calculate the force on a small sphere of $20\ \text{g}$ placed at $P$ and comment on why this experiment requires extraordinary sensitivity.