The law of universal gravitation

Two steel balls of $1\ \text{kg}$ each are placed so that their centres are $1\ \text{m}$ apart, as shown in the diagram.

Apply the law of universal gravitation, $F = G\cdot\frac{M\cdot m}{r^2}$, with $G = 6,67\cdot10^{-11}\ \text{N}\cdot\text{m}^2/\text{kg}^2$, and calculate the gravitational force attracting them. Does this seem like a large or small force? Comment on it.

In the law of universal gravitation, $F = G\cdot\frac{M\cdot m}{r^2}$:

1. a) State what each letter represents ($F$, $G$, $M$, $m$ and $r$) and its SI unit of measurement.
2. b) The value of $G$ is $6,67\cdot10^{-11}\ \text{N}\cdot\text{m}^2/\text{kg}^2$, a very small number. Explain the consequence of this for the force between objects of ordinary mass.

Two classmates, with masses of $60\ \text{kg}$ and $70\ \text{kg}$, are sitting such that their bodies are on average $1\ \text{m}$ apart. Calculate the gravitational force of attraction between them ($G = 6,67\cdot10^{-11}\ \text{N}\cdot\text{m}^2/\text{kg}^2$) and express it in standard form.

The Sun keeps the Earth in orbit by attracting it. Calculate this force using the following data: mass of the Sun $M_S = 1,99\cdot10^{30}\ \text{kg}$, mass of the Earth $M_T = 5,97\cdot10^{24}\ \text{kg}$, mean distance between their centres $r = 1,5\cdot10^{11}\ \text{m}$ and $G = 6,67\cdot10^{-11}\ \text{N}\cdot\text{m}^2/\text{kg}^2$. Then, compare the result with the Earth–Moon force, of about $2\cdot10^{20}\ \text{N}$: how many times larger is it?

Calculate the gravitational force of attraction between the Earth and the Moon, as shown in the diagram.

Given: mass of the Earth $M_T = 5,97\cdot10^{24}\ \text{kg}$, mass of the Moon $M_L = 7,35\cdot10^{22}\ \text{kg}$, mean distance between their centres $r = 3,84\cdot10^{8}\ \text{m}$ and $G = 6,67\cdot10^{-11}\ \text{N}\cdot\text{m}^2/\text{kg}^2$. Express the result in standard form, rounded to two decimal places.

Two twin oil tankers, each with a mass of $10^{6}\ \text{kg}$, are anchored with their centres of mass separated by $100\ \text{m}$.

1. a) Calculate the gravitational force attracting them ($G = 6,67\cdot10^{-11}\ \text{N}\cdot\text{m}^2/\text{kg}^2$).
2. b) Compare this force with that needed to drag a sheet of paper across a table, of the order of $0,01\ \text{N}$. Will the ships notice this attraction? Justify your answer.

Let us deduce the value of $g$ on the surface of the Earth from the law of universal gravitation.

1. a) Write the force exerted by the Earth on a body of mass $m$ in two ways: as $F = m\cdot g$ and as $F = G\cdot\frac{M\cdot m}{r^2}$.
2. b) Equate the two expressions, cancel the mass $m$ and rearrange for $g$.
3. c) Calculate its value with $M = 5,97\cdot10^{24}\ \text{kg}$, $r = 6,37\cdot10^{6}\ \text{m}$ and $G = 6,67\cdot10^{-11}\ \text{N}\cdot\text{m}^2/\text{kg}^2$. What well-known value do you obtain?

Two fixed masses, of $100\ \text{kg}$ and $400\ \text{kg}$, are $3\ \text{m}$ apart. We want to place a third mass on the line joining them, between the two, such that the net gravitational force on it is zero, as shown in the diagram.

1. a) If $x$ is the distance from the third mass to that of $100\ \text{kg}$, write down the equality between the two forces acting on it.
2. b) Solve the equation and calculate $x$ (hint: first simplify $G$ and the third mass, and then take square roots).
3. c) Does the result depend on the value of the third mass? What about $G$? Comment on this.