The Law of Universal Gravitation

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

Apply the law of universal gravitation, $F = G\times\frac{M\times m}{r^2}$, with $G = 6.67\times10^{-11}\ \text{N}\times\text{m}^2/\text{kg}^2$, 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, $F = G\times\frac{M\times m}{r^2}$:

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 $6.67\times10^{-11}\ \text{N}\times\text{m}^2/\text{kg}^2$, 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 ($G = 6.67\times10^{-11}\ \text{N}\times\text{m}^2/\text{kg}^2$) and express it in standard form.

Two asteroids attract each other with a force $F$.

1. a) If one of them had twice the mass, what would the new force be?
2. b) What if both of them had twice the mass?

Reason both answers using the formula for universal gravitation.

When you drop a pen, the Earth attracts it downwards; but, according to the law of universal gravitation, the pen also attracts the Earth with a force of equal magnitude.

1. a) Why do we only see the pen moving and not the Earth rising to meet it?
2. b) Reason this taking into account the masses of both bodies and the acceleration that the same force produces in each.

The Sun keeps the Earth in orbit by attracting it. Calculate this force using the following data: mass of the Sun $M_S = 1.99\times10^{30}\ \text{kg}$, mass of the Earth $M_T = 5.97\times10^{24}\ \text{kg}$, mean distance between their centers $r = 1.5\times10^{11}\ \text{m}$ and $G = 6.67\times10^{-11}\ \text{N}\times\text{m}^2/\text{kg}^2$. Then, compare the result with the Earth–Moon force, of about $2\times10^{20}\ \text{N}$: how many times larger is it?

Two bodies attract each other with a force $F$, as shown in the graph of force against distance.

Without changing the masses, we move them further apart until the distance between them is doubled. Looking at the formula $F = G\times\frac{M\times m}{r^2}$, what will the force be now? Justify your answer.

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

Given: mass of the Earth $M_T = 5.97\times10^{24}\ \text{kg}$, mass of the Moon $M_L = 7.35\times10^{22}\ \text{kg}$, mean distance between their centers $r = 3.84\times10^{8}\ \text{m}$ and $G = 6.67\times10^{-11}\ \text{N}\times\text{m}^2/\text{kg}^2$. Express the result in standard form, rounded to two decimal places.

At what distance from the center of the Earth is the gravitational force on a body reduced to half of that felt on the surface?

1. a) Set up the equality using the fact that the force decreases with $\frac{1}{r^2}$.
2. b) Rearrange for $r$ in terms of the Earth's radius $R_T$ (a $\sqrt{2}$ will appear).
3. c) Using $R_T = 6370\ \text{km}$, calculate at what height above the surface one must ascend, in kilometers. Do you think a commercial airplane, flying at a height of about $10\ \text{km}$, notices this decrease? Give reasons for your answer.

Calculate the force with which the Earth attracts a person of $50\ \text{kg}$ located on its surface, as shown in the diagram.

Given: mass of the Earth $M = 5.97\times10^{24}\ \text{kg}$, Earth's radius $r = 6.37\times10^{6}\ \text{m}$ and $G = 6.67\times10^{-11}\ \text{N}\times\text{m}^2/\text{kg}^2$. Compare the result with their weight, $P = m\times g$ with $g = 9.8\ \text{m/s}^2$: what do you observe?

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 $F = m\times g$ and as $F = G\times\frac{M\times m}{r^2}$.
2. b) Equate the two expressions, cancel the mass $m$ and rearrange for $g$.
3. c) Calculate its value with $M = 5.97\times10^{24}\ \text{kg}$, $r = 6.37\times10^{6}\ \text{m}$ and $G = 6.67\times10^{-11}\ \text{N}\times\text{m}^2/\text{kg}^2$. What well-known value do you obtain?

A rocky exoplanet has four times the mass of the Earth and twice the radius, as shown in the figure.

1. a) Using $g = \frac{G\times M}{R^2}$, reason by what factor gravity changes on its surface relative to Earth's gravity.
2. b) How much would a $60\ \text{kg}$ astronaut weigh there (take Earth's gravity as $9.8\ \text{m/s}^2$)?
3. c) What if the planet had four times the mass but the same radius as the Earth? Calculate the astronaut's new weight.

A geostationary satellite of $1000\ \text{kg}$ orbits at a distance of $r = 4.22\times10^{7}\ \text{m}$ from the center of the Earth, as shown in the diagram.

1. a) Calculate the force with which the Earth attracts it (mass of the Earth $M_T = 5.97\times10^{24}\ \text{kg}$, $G = 6.67\times10^{-11}\ \text{N}\times\text{m}^2/\text{kg}^2$).
2. b) Compare this force with the satellite's weight when it was on the surface, $P = m\times g$ with $g = 9.8\ \text{m/s}^2$: approximately how many times smaller is it?

Newton tested his law by comparing the fall of objects at the surface with the motion of the Moon.

1. a) Calculate the force that the Earth exerts on a body of $1\ \text{kg}$ on its surface ($r = 6.37\times10^{6}\ \text{m}$, $M_T = 5.97\times10^{24}\ \text{kg}$, $G = 6.67\times10^{-11}\ \text{N}\times\text{m}^2/\text{kg}^2$).
2. b) Calculate the force it would exert on that same body if it were at the Moon's distance ($r = 3.84\times10^{8}\ \text{m}$).
3. c) Verify that the second force is about $3600$ times smaller and relate that number to the ratio between the two distances, which is about $60$. What law is confirmed?

A book and the bookshelf on which it rests attract each other gravitationally, just like the Earth and the Moon. However, we never see the book move towards the shelf. Considering the value of $G$, the masses of the objects and the formula $F = G\times\frac{M\times m}{r^2}$, explain why we do not notice the gravitational attraction between everyday objects.

Two identical metal spheres, placed with their centers $2\ \text{m}$ apart, attract each other with a force of $6.67\times10^{-5}\ \text{N}$.

1. a) Write the law of universal gravitation letting $m$ be the mass of each sphere (they are both equal!).
2. b) Rearrange the formula for $m$ and calculate the mass of each sphere ($G = 6.67\times10^{-11}\ \text{N}\times\text{m}^2/\text{kg}^2$).

Calculate the gravitational force between a person of $80\ \text{kg}$ and a lorry of $5000\ \text{kg}$ parked $10\ \text{m}$ away ($G = 6.67\times10^{-11}\ \text{N}\times\text{m}^2/\text{kg}^2$). Then, calculate the weight of a mosquito of mass $2\ \text{mg}$ ($P = m\times g$, with $g = 9.8\ \text{m/s}^2$; be careful with the units) and compare both forces. What conclusion do you draw regarding the gravitational attraction between everyday objects?

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.