Intensity of the Gravitational Field 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: $G = 6.67\times10^{-11}\ \text{N}\times\text{m}^2/\text{kg}^2$, $M_T = 5.97\times10^{24}\ \text{kg}$, $R_T = 6.37\times10^{6}\ \text{m}$.
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 $M_M = 6.42\times10^{23}\ \text{kg}$, radius of Mars $R_M = 3.39\times10^{6}\ \text{m}$, $G = 6.67\times10^{-11}\ \text{N}\times\text{m}^2/\text{kg}^2$.
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 $M_L = 7.35\times10^{22}\ \text{kg}$, lunar radius $R_L = 1.74\times10^{6}\ \text{m}$, $G = 6.67\times10^{-11}\ \text{N}\times\text{m}^2/\text{kg}^2$.
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 of $10\ \text{kg}$ each are placed on a laboratory bench, separated by $2.0\ \text{m}$. Let $P$ be a point on the line joining them, located outside the segment, $1.0\ \text{m}$ from one mass and $3.0\ \text{m}$ from the other. The layout is shown in the figure.

1. a) Calculate the magnitude of the field created by each mass at $P$ and state its direction and sense ($G = 6.67\times10^{-11}\ \text{N}\times\text{m}^2/\text{kg}^2$).
2. b) As both fields have the same direction and sense, calculate the total field at $P$.

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 ($G = 6.67\times10^{-11}\ \text{N}\times\text{m}^2/\text{kg}^2$).
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 ($G = 6.67\times10^{-11}\ \text{N}\times\text{m}^2/\text{kg}^2$).
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 $x = 3.0\ \text{m}$.

The gravitational field of a point mass decreases with the square of the distance. The graph shows this dependence:

1. a) If at a distance $r$ from the center of a planet the field is equal to $g$, what is its value at a distance $2\times r$? Justify your answer using $g = G\times\frac{M}{r^2}$.
2. b) At what distance from the center of the Earth, expressed in Earth radii $R_T$, is the field equal to one-quarter of its value at the surface?
3. c) What height above the surface does this distance correspond to?

In a metrology laboratory, the goal is to measure the gravitational field created by a solid lead sphere of $100\ \text{kg}$.

1. a) Calculate the magnitude of the field at $0.50\ \text{m}$ from its center, treating the sphere as a point mass ($G = 6.67\times10^{-11}\ \text{N}\times\text{m}^2/\text{kg}^2$).
2. b) Express the result in scientific notation and compare its order of magnitude with that of the Earth's gravitational field ($9.8\ \text{m/s}^2$): what conclusion do you draw regarding the difficulty of the measurement?

Geostationary satellites orbit at a distance $r = 4.22\times10^{7}\ \text{m}$ from the center of the Earth, as shown in the figure.

1. a) Calculate the Earth's gravitational field at that distance ($G = 6.67\times10^{-11}\ \text{N}\times\text{m}^2/\text{kg}^2$, $M_T = 5.97\times10^{24}\ \text{kg}$).
2. b) How many times smaller is it than the field at the surface ($g_0 = 9.8\ \text{m/s}^2$)?
3. c) The field at a point does not depend on the mass placed at that point: explain what this means using the definition $\vec{g} = \frac{\vec{F}}{m}$.

Due to its huge mass, the Sun also creates a significant gravitational field at astronomical distances.

1. a) Calculate the magnitude of the Sun's gravitational field at the Earth's orbit ($M_S = 1.99\times10^{30}\ \text{kg}$, mean distance $d = 1.5\times10^{11}\ \text{m}$, $G = 6.67\times10^{-11}\ \text{N}\times\text{m}^2/\text{kg}^2$).
2. b) Compare the result with the field on the Earth's surface ($9.8\ \text{m/s}^2$): how many times smaller is it?
3. c) If the Sun pulls on us with this field, why doesn't it pull us off the Earth's surface? (Hint: the Earth and everything on it fall together towards the Sun.)

A geophysical prospecting gravimeter can detect variations in the gravitational field of up to $10^{-6}\ \text{m/s}^2$. It is suspected that there is a large body of water beneath a piece of land, which we approximate as a point mass of $10^{5}\ \text{kg}$ centred at a depth of $20\ \text{m}$. The diagram shows the layout:

1. a) Calculate the gravitational field created by the deposit on the surface, directly above it ($G = 6.67\times10^{-11}\ \text{N}\times\text{m}^2/\text{kg}^2$).
2. b) Could the gravimeter detect it? Justify your answer with the numbers.
3. c) State a practical application of this type of measurement.

An astronaut performs a spacewalk $10\ \text{m}$ from the center of the International Space Station, of mass $M_{ISS} = 4.2\times10^{5}\ \text{kg}$, which orbits at an altitude of $400\ \text{km}$ above the Earth's surface.

1. a) Calculate the Earth's gravitational field at the astronaut's location ($M_T = 5.97\times10^{24}\ \text{kg}$, $R_T = 6.37\times10^{6}\ \text{m}$, $G = 6.67\times10^{-11}\ \text{N}\times\text{m}^2/\text{kg}^2$).
2. b) Calculate the gravitational field created by the station itself at $10\ \text{m}$ from its center, treating it as a point mass.
3. c) Compare the orders of magnitude of both fields and explain why the astronaut does not "fall" toward the station despite its enormous size.
4. d) Which field dominates the astronaut's motion?

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 center of the Earth, write the condition for cancellation: $G\times\frac{M_T}{x^2} = G\times\frac{M_L}{(d - x)^2}$, where $d = 3.84\times10^{8}\ \text{m}$ is the Earth–Moon distance.
2. b) Simplify $G$, take square roots and solve for $x$ using $M_T = 5.97\times10^{24}\ \text{kg}$ and $M_L = 7.35\times10^{22}\ \text{kg}$ (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 $1.2\times10^{4}\ \text{kg}$ travels from the Earth to the Moon along the line joining their centers.

1. a) Calculate at what distance from the center of the Earth the total gravitational field (Earth plus Moon) vanishes. Data: $M_T = 5.97\times10^{24}\ \text{kg}$, $M_L = 7.35\times10^{22}\ \text{kg}$, $d = 3.84\times10^{8}\ \text{m}$, $G = 6.67\times10^{-11}\ \text{N}\times\text{m}^2/\text{kg}^2$; 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 toward the Moon: calculate the total field there (magnitude and direction) and reason whether the cancellation point is a position of stable or unstable equilibrium.

1. a) Calculate at what distance from the Earth, along the Sun–Earth line, the gravitational field of the Sun and that of the Earth cancel out. Data: $M_S = 1.99\times10^{30}\ \text{kg}$, $M_T = 5.97\times10^{24}\ \text{kg}$, $d = 1.5\times10^{11}\ \text{m}$; take $\sqrt{\frac{M_S}{M_T}} \approx 577$. Express the result in kilometers.
2. b) The Sun–Earth L1 Lagrange point, where solar observatories are located, is about $1.5\times10^{9}\ \text{m}$ from the Earth, much farther than predicted by your calculation. Explain what physical effect you have not taken into account (hint: the Earth and the observatory are not stationary, but orbiting around the Sun).

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 $P$ ($G = 6.67\times10^{-11}\ \text{N}\times\text{m}^2/\text{kg}^2$).
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 $L\times\sqrt{2}$). Data: $G = 6.67\times10^{-11}\ \text{N}\times\text{m}^2/\text{kg}^2$.
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?

Three equal point masses, $m = 150\ \text{kg}$, are at the vertices of an equilateral triangle of side length $L = 3.0\ \text{m}$. The centroid is at a distance $\frac{L}{\sqrt{3}}$ from each vertex. The figure shows the layout:

1. a) Reason, using symmetry and the addition of three equal vectors forming $120^\circ$ with each other, that the total field at the centroid is zero.
2. b) One of the masses is removed: applying superposition, show that the field at the centroid is equal and opposite to the field created by the removed mass, and calculate its magnitude ($G = 6.67\times10^{-11}\ \text{N}\times\text{m}^2/\text{kg}^2$).
3. c) Calculate the force that a mass of $20\ \text{kg}$ placed at the centroid would experience, and state its direction and sense.

A binary star system consists of two stars of the same mass, $M = 2.0\times10^{30}\ \text{kg}$, separated by $d = 2.0\times10^{11}\ \text{m}$, as shown in the figure.

1. a) Reason that the total gravitational field at the midpoint of the line segment joining the stars is zero.
2. b) Calculate the total field at a point on the perpendicular bisector of that segment located at $h = 2.0\times10^{11}\ \text{m}$ from the midpoint: draw the vectors in your notebook, reason which components cancel out, and calculate the result ($G = 6.67\times10^{-11}\ \text{N}\times\text{m}^2/\text{kg}^2$).
3. c) If a small mass probe is released from rest at that point, describe qualitatively how it begins to move.

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

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 $P$ ($G = 6.67\times10^{-11}\ \text{N}\times\text{m}^2/\text{kg}^2$).
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.

The origin of the tides lies in the fact that the Moon's gravitational field is non-uniform across the Earth. The diagram shows the geometry of the problem:

1. a) Calculate the magnitude of the lunar field at the point on the Earth's surface closest to the Moon (its distance to the Moon is $d - R_T$).
2. b) Calculate the magnitude of the lunar field at the point on the surface furthest from the Moon (distance $d + R_T$).
3. c) Calculate the difference between the two and explain why this difference, although small, "stretches" the oceans and produces two high-tide regions. Data: $M_L = 7.35\times10^{22}\ \text{kg}$, $d = 3.84\times10^{8}\ \text{m}$, $R_T = 6.37\times10^{6}\ \text{m}$, $G = 6.67\times10^{-11}\ \text{N}\times\text{m}^2/\text{kg}^2$.