The gravitational constant determines the strength of gravitational attraction. Changing it opens up a wide range of possible simulations that explore how gravity shapes everyday life, planets, stars and the universe.

The gravitational constant G appears in Newton’s law of universal gravitation as well as Einstein’s general theory of relativity. In Newton’s law of gravitation G proportionalizes the attractive gravitational force between two bodies,

where m₁  and m₂  are the masses of the point-like bodies and r is the distance between them. Both bodies act with an equal and opposite force (F₁ and F₂) towards each other according to Newton’s third law.

In this chapter we look at what would happen if the value of G is either decreased or increased. The starting assumption is that everything else, such as the mass of the Earth, will remain the same; the only entity changing is the gravitational constant. How would the world look, if we adjusted the fine-settings just a bit?

There is no clear border between the atmosphere and empty space; the atmosphere does not stop abruptly. However we can think of the borderline to be located at a distance from the Earth, where the gravitational force of the Earth is equal to the force that is creating Earth’s centripetal acceleration

where R is the distance from the centre of the Earth to the molecule and ω is Earth’s angular velocity. If we observe a single molecule in the atmosphere that is in a rotating motion with the Earth, assuming that the molecule holds its place in relation to the Earth, the centripetal force can be expressed with Newton’s second law

where m₁ is the molecule’s mass and aR is the centripetal acceleration. This can be written further into

From this we can solve R to be

This manner of an approach would seem to say that the stronger the gravity, i.e. the larger the gravitational constant G, the distance from the centre of the Earth to the edge of the atmosphere would increase. Why would the atmosphere widen if the gravitational force pulled the molecules tighter towards the Earth. The answer lies in the starting assumption; unlimited molecules. The directly proportional nature of G and R means that if there was an unlimited amount of molecules around Earth in space, molecules from further away would be lured in to Earth’s atmosphere. The Earth would be able to attract a wider cloud of molecules to follow itself.

In reality this approach is flawed if we want to look what happens to a specific molecule and follow its path as G changes. Defining the end of the atmosphere as an altitude where the atmospheric pressure is defined as some certain value helps to observe the changes in the thickness of the atmosphere as G varies.

One way to do this is to look at the Kármán line which is one arbitrary way to define the edge of space, and so the end of the atmosphere. It is based on the altitude where an aeroplane has to reach orbital velocity to generate sufficient lift. The altitude of the Kármán line depends from organisation to organisation, but NASA for example defines the line to be at 80 kilometres above sea level. The interest in defining such an altitude lies in the differentiation between aeronautics and astronautics; a person who has flown above the Kármán line is defined as an astronaut. The atmospheric pressure at the Kármán line is roughly 0.3Pa. 

Earth’s atmosphere is not smooth. The temperature and its lapse rate change depending on altitude. The atmosphere can be divided into seven layers for calculative purposes. The first of those seven begins from sea level and goes up to 11000m. In each layer the temperature and its lapse rate are constant. In our current atmosphere pressure as a function of altitude through all of the seven layers, up to 86 kilometres, is shown in the figure below.

The atmospheric pressure on Earth’s surface depends on the gravitational constant. As gravity is directly proportional to G, when G increases so does the pressure on the surface, as the Earth is pulling air molecules tighter against itself.The surface pressure can be approximated with the equation of hydrostatic equilibrium, that states

where P is atmospheric pressure,r is a distance from the centre of the Earth, M(r) is the mass that is enclosed in the r-radius sphere and ρ is in this case the mean density of Earth’s atmosphere which is

where h is altitude and R is the radius of the Earth. By considering dP/dr as the pressure difference between the surface of the Earth and the edge of the atmosphere that is located at an altitude h we get

Assuming that the pressure at the end of the atmosphere is negligible compared to the surface pressure we get

for the atmospheric pressure on Earth’s surface.

Using 5.9772×10²⁴ kg as Earth’s mass, 5.1480×10¹⁸  kg as the mean of Earth’s atmosphere’s mass and setting h at the Kármán line at 80 kilometres we get an approximation for the surface pressure of about 95kPa. As the mass of the atmosphere varies considerably due to water vapour the estimate is rough - but still quite close to the measured value of approximately 100kPa. However, from the above we can see that P(R)∝G. In other words, if the gravitational constant becomes n×G_(initial) where n >0, the new surface pressure is simply n×P(R)_(initial) as shown in the figure below.

The same approach applies to the atmospheric density on Earth’s surface. Assuming that the temperature does not change, the equation of state can be described by the ideal gas law

where R is the universal gas constant, M_m is the molar mass of air and T is temperature. As calculated above, the atmospheric pressure P(R) on Earth’s surface has the gravitational constant G as a factor. As the atmospheric density on Earth’s surface is directly proportional to the surface pressure it also has G as a factor in it. As a result, if the gravitational constant becomes n×Gᵢₙᵢₜᵢₐₗ where n >0, the new atmospheric density on Earth’s surface is simply n×ρ(P(R))ᵢₙᵢₜᵢₐₗ. At 101325Pa and 15°C the density of air is approximately 1.2kg/m³. 

In our current atmosphere the atmospheric density as a function of altitude through all of the seven layers, up to 86 kilometres is shown in the figure below.

Finding the pressure - and density - distribution of the atmosphere after G has changed can be done with the barometric formula for pressure or density that - for the first layer of the atmosphere - states the pressure as a function of altitude and the density as a function of altitude to be

where Pᵣ is the pressure on the surface of the Earth as explained before; ρᵣ is the atmospheric density on Earth’s surface; R is the universal gas constant; M is the molar mass of Earth’s air; T is the standard temperature (K) and L is the standard temperature lapse rate (K/m), in other words the rate at which the temperature in Earth’s atmosphere changes with altitude.

In these equations the gravitational acceleration g is treated as a constant, but the gravitational constant G appears only in the definition of the gravitational acceleration g, as

where R Earth is Earth’s radius and h is altitude. Therefore, we treat g as a variable.

Increasing G would theoretically allow Earth to obtain molecules from a distance. For example a sudden doubling in the value of G would lead to an obtain range of ≈1.26×Rᵢₙᵢₜᵢₐₗ. More generally, if the gravitational constant becomes n times its initial value, Earth’s obtain radius becomes the third root of n times R_(initial), where R_(initial) is the initial distance to the edge of the atmosphere that can be calculated above.

Observing the existing atmosphere and the new structure it gains from the change in the value of the gravitational constant, we can for example find the new altitude of the Kármán line, when defined as an altitude where the atmospheric pressure equals ≈0.3Pa. Since the atmosphere is layered for calculative purposes, for the sake of accuracy it is sensible to only observe the first layer that reaches up to 11 kilometres.

G would have to increase tenfold for the Kármán line to fall just at the end of the first atmospheric layer, at approximately 11 kilometres. This means that a person who has flown above this limit would be defined as an astronaut by some organisations. Similarly, the atmospheric density as a function of altitude with different values for the gravitational constant is shown in the figure below. Using the equation where the temperature lapse rate does not equal zero. Therefore, we are approximating the above expression for ρ(P) with the barometric equation for a situation where the temperature lapse rate equals zero. For the first layer it states

If we were to use the original expression instead, with the temperature lapse rate not being zero, the density of the atmosphere would go up as altitude increased with sufficiently small values of G. This situation is shown in the figure below.

When the gravitational constant has a value of 0.1G the density of the atmosphere increases with altitude. At an altitude of 1 meter the density is approximately 0.12kg /m³ and at the end of the first atmospheric layer, at 10999 meters the density is approximately 0.14kg/m³. Below is a logarithmic representation of the figure.

Earth is in a rotational motion with an angular velocity of ω around its axis. Let’s observe a body fixed on Earth’s surface.It is also in a rotational motion with the same angular velocity and hereby subjected to a centripetal acceleration. The centripetal force equals

Angular velocity can be expressed as

where L is the angular momentum of Earth. This quantity is conserved unless the system is acted upon by an external torque. Torque is defined to be

where θ is the angle between the moment arm R and the force vector in question. There can be no angular acceleration,i.e. change in angular velocity without torque.

Theoretically the only force on the surface of the spinning Earth is Earth’s own gravitational force that is directed vertically downward, toward Earth’s centre. This means that it cannot create a torque since the angle θ is zero. As a result, a change in the value of the gravitational constant which changes the strength of the gravitational force does not affect Earth’s rotation. As stated above, in a system with no torque angular momentum is conserved. From this we get a relation (assuming that the Earth’s mass remains the same)

The only way for the new ω to differ from the old ω is ifthe ratio of the old to the new R² is not 1. In other words, only by changing the Earth’s radius can its angular velocity be altered.

The Moon’s rotation around our planet can be expressed as

where gravity equals the centripetal force.

By substituting ω with  L/(mR²), where the angular momentum L is conserved we get for the orbital radius the following relation:

The orbital radius is inversely proportional to the gravitational constant. As G increases the gravitational force becomes stronger and the Moon will fall to a closer orbit. Similarly, if G decreases the Moon’s orbital radius will grow.

The orbits of the Earth and the Moon can be solved as a two-body problem in barycentric coordinates, in this case using Python. A sudden change in the value of G results in a high eccentricity orbit for the Moon. A gradual change lets the new orbit to stabilise and the orbit retains its shape. One month is considered to equal 27.3 days, that contain 86400 seconds.

Earth’s own radius would change as G changes. Increasing the gravitational constant makes gravity stronger, pulling everything tighter together. Theoretically, Earth’s radius would decrease and our planet would become denser. Since Earth’s surface would be closer to the centre of its mass, we would experience a greater acceleration

due to a larger G and a smaller Earth's radius.

In reality, however, it seems that the Earth’s radius would remain the same in relatively small changes in G’s value, due to Earth’s crust and its stiffness. Just as a glass sphere will not immediately break if a force is applied evenly over its surface, or the pressure inside it is increased (which would be somewhat analogous to decreasing the value of G perhaps), the Earth will keep its shape and radius to some borderline values of G. If G is increased it would seem intuitive that once the borderline value has been reached, Earth’s crust would crumble into pieces and the planet would shrink to a denser object. Before this happened, when the radius is still the same, the gravitational force on Earth would increase. As a result the acceleration on Earth’s surface would become larger; things would fall down faster, objects would weigh more, moving around - for example walking - would require more energy if the movement had a vertical component and the air pressure would increase as well. The temperature inside Earth’s crust would perhaps rise if the pressure inside Earth increased enough. 

If G is decreased the inner layers of the Earth would seek to expand in volume as the gravitational force keeping them under pressure would diminish. This would perhaps result in volcanic activity as the molten layers of Earth erupted. Also acceleration on Earth’s surface would decrease, and the opposites would happen than in the situation where G is increased.

Tidal phenomena are distortions in the shape of an object caused by the gravitational force of another celestial body. On Earth the most known and visible example is the phenomenon of oceanic tides, the rising and falling of the water surface that is most noticeable near shores as the shoreline moves back and forth. In a sense every celestial object has an effect on the tides on Earth through their gravitational field. However, only the effects of the Sun and the Moon are significant. The ratio of the lunar and solar tide-generating forces on the Earth is approximately 2.2, i.e. the Moon has a larger effect.Because of this, for simplicity, we will only examine the tidal phenomena caused by the lunar tide-generating forces. In reality the distribution of continents, the variation in the depth of the oceans and the dynamical response of the oceans to these tide-generating forces cause an immensely complicated system of tides that is nearly impossible to predict with full accuracy. However, the forces in question are well known and we can observe their effect on a simplified model of planet Earth. This will bring clarity to the physics of the tides and allow us to obtain a sufficient understanding on what would happen if the value of the gravitational constant was altered. In the following segments we simplify the oceans of the Earth to a uniform shell of water covering the whole planet with an equal depth. Also we fix the Moon to an equatorial orbit and observe a non-rotating (non-spinning) Earth.

The Earth and the Moon orbit around their common centre of mass in revolutions without rotation. In other words, their movement is translational - the axis of a non-rotating reference frame, with its origin at the centre of the Earth, are fixed toward immovable stars. For the Earth, the gravitational attraction of the Moon produces an acceleration that is independent of the orbital velocity around the common centre of mass. It is simply the acceleration of free fall and it would be the same if Earth were to freely fall toward the Moon. The magnitude and direction of this free fall acceleration a₀ is the same in the same gravitational field. When observing a body on Earth in the reference frame mentioned above, a pseudo force of inertia acts on it:

This force is the same to all objects with mass m in the proximity of planet Earth; the force field is uniform. On the other hand, the gravitational field of the Moon that is causing the free fall acceleration is not uniform. In the centre of the Earth this pseudo force of inertia and the gravitational attraction of the Moon cancel out. On the Earth’s surface nearest to the Moon the gravitational force is larger than the pseudo force of inertia and so the sum of these forces is directed toward the Moon, away from the centre of the Earth. On the other side of the Earth the pseudo force of inertia is in turn larger than the Moon’s gravitational attraction, so the net force is directed away from the Moon and the centre of the Earth. In conclusion, the sum of the pseudo force of inertia and the gravitational attraction of the Moon on Earth’s surface, on a line connecting the centres of these two celestial objects, is always directed away from the centre of the Earth. This combined effect of pseudo force of inertia and the gravitational force is what we call the tidal force.

On Earth’s surface at the points where the Moon is at the zenith or at the nadir, the tidal force is directed vertically upward. The points on Earth where the Moon is at the horizon experience a tidal force that is directed vertically downward. The ocean shell of planet Earth is in a way squeezed symmetrically along the Earth-Moon line. This results in the so called tidal bulges that are located on the Earth-Moon line with one of them at the point where the Moon is at the zenith and the other one at the point where the Moon is at the nadir.

The horizontal and vertical components of the tidal force can be simplified (Butikov, The Physics of the Oceanic Tides) to be

where R is the distance between the Earth and the Moon, r is the distance from the centre of the Earth to the body in question (a drop of water for example) and θ is the angle between the direction to the Moon and to the body in question as seen from the centre of the Earth. It is worth noting that the horizontal component is much more impactful than the vertical one; it is responsible for shifting the waters around the globe.

With the starting assumptions and simplifications that the planet Earth is not rotating and so the tide-generating forces are nearly time-independent we can examine the distortion of the oceans due to these tidal forces. The static distortion has a spheroidal shape that is given by

where r₀ is the distance from the centre of the Earth without any tidal effects and 2a is the difference between the smallest and the greatest value of r(θ). The value 2a can be expressed as a relation between the horizontal tidal force and the gravitational force as (Butikov, The Physics of the Oceanic Tides)

From the above we get 2a≈0.5353m for the difference between the static minimal and maximal levels of the ocean surface due to lunar tide-generating forces.

We see that the static distortion of the ocean surface is not directly affected by the gravitational constant. Indeed, increasing G would increase the Earth’s gravitational attraction and so keep the oceans in a stronger hold. However, the Moon’s gravitational attraction would also increase in the same fashion so the overall effect would cancel out.

The value of 2a is inversely proportional to the cube of the Earth-Moon distance. This is indeed affected by the changing of the gravitational constant, as the Moon would obtain a new orbit if the mutual gravitational attraction was altered. A sudden change in the gravitational constant would result in an elliptical orbit for the Moon. The eccentricity of the new orbit is proportional to the magnitude and the rate of the change. If G changes gradually the shape of the orbit will remain quite similar.

The Moon’s average distance to the Earth will decrease when the gravitational constant increases since the Earth’s gravitational force will become stronger and pull the Moon to a closer orbit. If the orbit keeps its shape and only the radius of the orbit changes, the difference in height between a low tide and a high tide represented by 2a will in this situation become greater. If the gravitational constant was decreased and the Moon fell to a further orbit all the while keeping the shape of its initial orbit, the value 2a would decrease. The difference between high and low tide would become smaller.

If the value of the gravitational constant was changed suddenly the Moon would acquire an orbit of high eccentricity. In this situation the magnitude of the tidal force would be time-dependent; as the Moon is at a point in its orbit where it is closest to the Earth the difference in height between a low tide and a high tide would be greater than when the Moon is at the furthest point in its orbit.