Equations Introduced and Used in this Topic:

• [latex]F_g=\dfrac{Gm_1m_2}{d^2}[/latex]
• [latex]a_g \text{ or }g = \dfrac{Gm}{{d}^2}[/latex]
• [latex]\Delta V=-\dfrac{Gm}{d}[/latex]

G = 6.67408(31) × 10⁻¹¹ Nm²/kg² (round to needed accuracy)

Where…

• [latex]F_g[/latex] is the Gravitational Force of attraction, measured in newtons (N)
• [latex]G[/latex] is Newton’s Gravitational Constant, currently estimated to be 6.67408(31) × 10⁻¹¹ Nm²/kg²
• [latex]m_1[/latex] & [latex]m_2[/latex] are the Masses of the two interacting bodies, measured in Kilograms (kg)
• [latex]a_g[/latex] or [latex]g[/latex] is the Gravitational Field Strength of a body, measured in metres/second squared (m/s2) or newtons per Kilogram (N/kg)
• [latex]d[/latex] is the Distance away from the Mass Centre of a Body (gravitational field) or the Distance between Mass Centres of Two Bodies (gravitational fields), measured in metres (m)
• [latex]∆V[/latex] is the potential difference measured in J/kg

• Solar Mass (m_s) is the mass of the Sun. 1 Solar mass = 1.98855 ± 0.00025 × 10³⁰ kg
• Astronomical Unit (AU) is the distance from Earth to the Sun. 1 AU = 1.49597870700 × 10¹¹ m

17.1 Gravitational Forces

Equations Introduced or Used for this Section:

[latex]F_{g}=\dfrac{Gm_1m_2}{{d}^2}[/latex]

Conceptions of Gravity and Astronomical Motion

Recorded concepts of gravity in Antiquity centre around the Greek philosopher Aristotle (384-322 BCE). Aristotle’s concept of gravity emerged from his understanding of natural motion in that all matter will rise or fall to its natural place in the universe. He contrasted this with forced or violent motion, where an object could be made to move until the agent causing it to move stopped.  In Aristotle’s paradigm, gravity was a manifestation of objects moving to simply reach their natural place in the universe.

Aryabhata (476-550) held the view that the Earth rotates daily on its axis, but held onto the heliocentric model of the universe rotating around the Earth, as taught by the earlier pre-Ptolomaic Greek astronomers. While a great deal of Aryabhata’s work is lost, remnants indicate that he might have believed the planets move in elliptical orbits, a view that took nearly a thousand years to resurface in the works of Nicolaus Copernicus (1473-1543). There is also evidence that Aryabhata considered gravity to be an attractive force that allowed people to stand upright no matter where they were positioned on the Earth.

Galileo’s (1564-1642) was the first to record counter arguments to Aristotelian views on the movement of bodies to their natural place in the universe, through his experiments with the acceleration of falling bodies. Galileo’s concepts of acceleration were later confirmed by the works of Francisco Maria Grimaldi (1618-1663) and Giovanni Battista Riccioli (1589-1671). It was Riccioli  who, through his analysis of free-fall motion in his attempts to find errors in Galileo’s theories, ended up finding the first measure of the acceleration of gravity. He estimated this to be 9.36 ± 0.22 m/s², which is within a 5% error of today’s accepted value of 9.8 m/s².

In working to quantify Robert Hooke’s  (1635-1703) suggestion that gravitational force was an inverse square law relation , Isaac Newton (1643-1727) came up with the proportionality relation:

[latex]{F}_g∝ \dfrac{m_1{m}_2}{{d}^2}[/latex]

The first measured value of the constant G needed for Newton’s equation came from the 1797 experiments by Henry Cavendish (1731-1810) . This was accomplished by Cavendish’s measure of the density of the Earth, which he measured to be 5.48 times that of water. Cavendish’s work led other natural philosophers to estimate a value for G to be 6.754 × 10⁻¹¹ N-m²/kg² close to the current accepted value of 6.67428 × 10⁻¹¹ N-m²/kg². The Cavendish experiments are considered to be highlights of accuracy in laboratory science; his work is taught as representing a gold standard in experimental research.

Shown below is a sketch of the torsion apparatus that Cavendish used to measure the gravitational constant G.

17.2 Gravitational Field Strength

Equations Introduced or Used for this Section:

[latex]a_g \text{ or }g=\dfrac{Gm}{d^2}[/latex]

In Newtonian mechanics or Classical Mechanics (non-relativistic) an equation quantifying gravitational strength of any body can be found by equating Newton’s Second Law and Newton’s Law of Gravitational Force. This can be shown as follows:

[latex]F_{\text{net}}= ma\text{ or }F_g = mg\text{ is equated to }F_{g}=\dfrac{Gm_1{m}_2}{{d}^2}[/latex]

This means that …

[latex]mg=\dfrac{Gm_1{m}_2}{{d}^2}[/latex]

cancelling out the common mass leaves us with …

[latex]g=\dfrac{Gm_1}{{d}^2}[/latex]

Applying the conventional negative direction to this derivation we are left with

[latex]g=-\dfrac{Gm_1}{{d}^2}[/latex]

In most classical mechanics questions this negative value is left out and the strength of gravity is just given as a scalar value.

The units of gravitational field strength can be either m/s² or N/kg.   This equation relates to the gravity at any distance away from a point mass and is an inverse square law relationship.

[latex]g ∝ \dfrac{1}{{d}^2}[/latex]

When we look at the gravitational field strength as we move towards the centre of the mass, the gravitational field strength drops inversely to zero. Beneath the planet’s surface gravity is an inverse linear relationship.

[latex]g ∝ \dfrac{1}{{d}}[/latex]

17.3 Gravitational Potential

Equations Introduced or Used for this Section:

[latex]ΔV =-\dfrac{Gm_1}{d}[/latex]

In classical mechanics, the gravitational potential at any location equals the work per unit of mass needed and moves the object from a fixed reference position to a final position within or outside a gravitational field. Typically, the gravitational potential refers to some final position infinitely far away from the gravitational source. As a result, at an infinite distance away the potential energy of any object in relation to its source position is zero. (Epf = 0). Also note that the movement that changes an object’s potential energy must be either toward or away from the gravitational source.

If we look at the work needed to move an object away from a gravitational source, we start by using the conservation of mechanical energy … Specifically:

[latex]W = ∆\text{Energy}[/latex]… Since the [latex]∆ E[/latex] is a change in potential energy… [latex]W = E_{pf} −E_{pi}[/latex]

This can be simplified for cases where the object is moved to an infinite distance away to

[latex]W = - E_{pi}\text{ since }E_{pf} = 0[/latex]

We can expand [latex]W = - E_{pi}[/latex] to [latex]W = - m g h_i[/latex] and replace gravity with [latex]g= - \dfrac{Gm_1}{d}[/latex]

This yields: [latex]W= - m\dfrac{Gm_1}{d}{h}_{i}[/latex]… And since h_i and d₂ are measuring the same distance, we can cancel common terms.

This leaves us with [latex]W=-\dfrac{Gm_1{m}_2}{d}[/latex]

The derivation of gravitational potential is almost complete. What remains is to isolate the second mass m₂.

Isolating m₂ from the main equation leaves us with: [latex]W=-\dfrac{Gm_1}{d}({m}_2)[/latex]

The symbol we will use for gravitational potential is the same as the one we use for electric potential. Gravitational potential (∆V) where [latex]\Delta V=-\dfrac{Gm_1}{d}[/latex]

In this case, the work done (joules) to move an object infinitely away from a measurable gravitational field is found by the equation:

[latex]W=-\dfrac{Gm_1}{d}({m}_2)[/latex]                    or                   [latex]V=−\Delta V (m_2)[/latex]

Likewise the gravitational potential energy of an object removed from the influence of a planetary or sun’s gravitational field is also found by:

[latex]\Delta E_p=-\dfrac{Gm_1}{d}({m}_2)[/latex]                  or                 [latex]\Delta E_p=-\Delta V (m_2)[/latex]

How do we understand the negative value that shows up in these equations?

• As you move away from a gravitational field, you will need energy to escape. You can consider this energy to be lost.
• As you move into a gravitational field, you will be gaining energy, which is added to your mass.

Equation Summary: To move any mass away from a gravitational source, use:

[latex]W=-\dfrac{Gm_1{m}_2}{d}[/latex]                  or                    [latex]\Delta E_p=-\dfrac{Gm_1{m}_2}{d}[/latex]

Using ∆V = [latex]-\dfrac{Gm_1}{d}[/latex] the above equations shorten to:

[latex]W=-\Delta Vm_2[/latex]                          or                  [latex]\Delta E_p=-\Delta Vm_2[/latex]

17.4.1 Earth-Moon Gravity Neutral Point

Between the Earth and the Moon, there exists a “fluctuating” position where the gravitational fields of the Earth and Moon balance each other out. This position is defined as the Earth-Moon Gravity Neutral Point.  Described by an equation, this position can be found from

[latex]g_{\text{ Earth}}= g_{\text{ Moon}}[/latex]

where [latex]g=\dfrac{GM}{{d}^2}[/latex]

Technically, this position is where the gravitational fields from both the Moon and the Earth balance each other out, meaning that any object in this position having no velocity relative to the Earth and Moon should stay in that position without being pulled to the Earth or the Moon. Should any spaceship reach this position headed to Earth or the Moon, it would then be able to fall to either side of this position without using any more fuel for thrust. This can be looked at as a moving vehicle at the top of a hill … If the vehicle is moving in any direction down the hill when at the top, then the vehicle will be able to roll down hill without using any fuel. If the vehicle is stopped at the top of the hill, then it should be able to remain stationary unless some external unbalanced force acts upon it, which would then act to nudge the vehicle downhill in one of the two directions. This position is termed fluctuating because of a number of factors which are described by NASA’s David Ellis.

The gravity neutral position is visually shown by mapping the variable strength of the Earth’s gravitational field as one heads to the Moon, overlaid by the variable strength of the Moon’s gravitational field as one heads back to the Earth. The intersection of these two gravitational fields’ lines marks the approximate position. Due to fluctuations in the location of this position, any spaceship parked there would need corrections by booster rockets to remain in the region of neutral gravity.

Exercise Answers

17.1 Gravitational Force

1.  9.82 N
2. 9.3 × 10⁻¹⁰ N
3. 5000 m or 5.00 km
4. 3.63 × 10⁻⁴⁷ N
5. 5.52 × 10²² N
6. 1.91 × 10¹⁸ N
7. Earth-Moon: 2.02 × 10²⁰ N,
   Sun-Moon: 4.39 × 10²⁰ N

17.2 Gravitational Field Strength

1. 1.2 × 10⁻⁸ m/s²
2. 9.01 m/s² 
3. 272 m/s²
4. 8.96 m/s²
5. Moon: 2.70 × 10⁻³ m/s²
   Earth: 3.32 × 10⁻⁵ m/s²
6. 9.01 × 10⁶ m from the Earth’s centre or 2650 km above the surface
7. 2.45 m/s² or 1/4 g’s
8. 1.25 × 10⁶ m  or 1250 km
9. 3.49 × 10⁵ m

17.3 Gravitational Potential

1. [latex]\Delta V[/latex] = − 5.97 × 10⁴ J/kg
2. [latex]W[/latex] = − 7.77 × 10¹⁰ J
3. [latex]\Delta V[/latex] = − 2.82 × 10⁶ J/kg
4. [latex]\Delta V[/latex] = − 8.87 × 10⁸ J/kg

17.4.1 Earth-Moon Gravity Neutral Point

1. Distance from the Earth = 3.46 × 10⁸ m & Distance from the Moon = 3.84 × 10⁷ m

Media Attributions

• “Cavendish Experiment” by Henry Cavendish is in the public domain.
• “Gravity Potential” by AllenMcC. is licensed under a CC BY-SA 3.0 licence.

• “Earth moon gravitational potential” by MikeRun is licensed under a CC BY-SA 3.0 licence.