1.5. Forces acting in a gravitational field Notes | Leaving Cert Physics

Universal Gravitation and Kepler's Third Law

Newton's law of universal gravitation states that the attractive force between any two point masses is directly proportional to the product of their masses and inversely proportional to the square of the distance between their centres:

F = \frac{G m_1 m_2}{r^2}

where $G = 6.67 \times 10^{-11} \text{ N m}^2 \text{ kg}^{-2}$ is the universal gravitational constant, and $r$ is the separation between their centres.

When an object of mass $m$ orbits a central body of mass $M$ in a circular path of radius $R$ , gravity supplies the required inward centripetal force:

F_g = F_c \implies \frac{GMm}{R^2} = \frac{mv^2}{R}

Dividing both sides by $m$ and multiplying by $R$ yields the square of the orbital speed:

v^2 = \frac{GM}{R} \implies v = \sqrt{\frac{GM}{R}}

The mass of the orbiting satellite $m$ cancels completely. Orbital speed depends only on the mass of the central body $M$ and the orbital radius $R$ .

Deriving Kepler's Third Law

To express this relationship in terms of orbital period $T$ , we use the linear speed expression $v = \frac{2\pi R}{T}$ :

v^2 = \frac{4\pi^2 R^2}{T^2}

Equating this to our gravitational orbital speed expression:

\frac{4\pi^2 R^2}{T^2} = \frac{GM}{R}

Cross-multiplying to isolate $T^2$ gives:

T^2 = \left(\frac{4\pi^2}{GM}\right) R^3

Because $4$ , $\pi$ , $G$ , and $M$ are constant for any chosen central body, Kepler's third law shows that the square of the orbital period is directly proportional to the cube of the orbital radius ( $T^2 \propto R^3$ ).

Verifying Kepler's Third Law Using Secondary Data

We can verify this law by examining planetary data for bodies orbiting the Sun:

Planet	Orbital Radius $R$ ( $10^{11}\text{ m}$ )	Period $T$ ( $10^7\text{ s}$ )	$\frac{T^2}{R^3}$ ( $10^{-19}\text{ s}^2\text{ m}^{-3}$ )
Venus	$1.08$	$1.94$	$2.99$
Earth	$1.50$	$3.16$	$2.96$
Mars	$2.28$	$5.94$	$2.98$

The ratio $\frac{T^2}{R^3}$ remains constant within experimental uncertainty. A graph of $T^2$ on the y-axis against $R^3$ on the x-axis gives a straight line passing through the origin. The slope of this line is $\frac{4\pi^2}{GM}$ , from which the mass of the central body can be calculated directly:

M = \frac{4\pi^2}{G \times \text{slope}}

A schematic graph of period squared against orbital radius cubed is a straight line through the origin, with slope 4π² divided by GM.

Gravitational Field Strength and Evidence from the Earth–Moon System

The gravitational field strength $g$ at any distance $r$ from the centre of a celestial body of mass $M$ is defined as the gravitational force per unit mass:

g = \frac{GM}{r^2}

When an object is above the surface of a planet of radius $R_p$ at an altitude $h$ , the total distance is $r = R_p + h$ . For instance, for the International Space Station orbiting at an altitude of $h = 400\text{ km}$ ( $4.0 \times 10^5\text{ m}$ ) around Earth ( $M = 6.0 \times 10^{24}\text{ kg}$ , $R_E = 6.4 \times 10^6\text{ m}$ ):

r = 6.4 \times 10^6 + 4.0 \times 10^5 = 6.8 \times 10^6\text{ m}

g = \frac{(6.67 \times 10^{-11})(6.0 \times 10^{24})}{(6.8 \times 10^6)^2} \approx 8.7\text{ m s}^{-2}

This is roughly $89\%$ of the value at Earth's surface. Gravity in low Earth orbit is definitely not zero.

Don't confuse orbital speed $v = \sqrt{\frac{GM}{R}}$ with escape velocity $v_e = \sqrt{\frac{2GM}{R}}$ , which is the minimum launch speed required at the surface for an unpowered projectile to escape a celestial body's gravitational field entirely. For Earth, $v_e \approx 1.1 \times 10^4\text{ m s}^{-1}$ (about $11\text{ km s}^{-1}$ ), which is $\sqrt{2}$ times the circular orbital speed at the surface.

Does Gravity Provide the Centripetal Force? The Earth–Moon Check

To test whether Newton's universal gravitation actually accounts for lunar motion, we compare the required centripetal force against the gravitational pull using astronomical data:

Distance between centres: $r = 3.84 \times 10^8\text{ m}$
Orbital period of the Moon: $T = 27.3\text{ days} = 27.3 \times 24 \times 3600 = 2.359 \times 10^6\text{ s}$
Mass of Earth: $M = 5.97 \times 10^{24}\text{ kg}$
Mass of Moon: $m = 7.35 \times 10^{22}\text{ kg}$

First, find the Moon's angular velocity:

\omega = \frac{2\pi}{T} = \frac{2\pi}{2.359 \times 10^6} \approx 2.66 \times 10^{-6}\text{ rad s}^{-1}

Second, calculate the centripetal force required to hold the Moon in this orbit:

F_c = mr\omega^2 = (7.35 \times 10^{22})(3.84 \times 10^8)(2.66 \times 10^{-6})^2 \approx 2.00 \times 10^{20}\text{ N}

Third, calculate the gravitational attraction between Earth and the Moon:

F_g = \frac{GMm}{r^2} = \frac{(6.67 \times 10^{-11})(5.97 \times 10^{24})(7.35 \times 10^{22})}{(3.84 \times 10^8)^2} \approx 1.98 \times 10^{20}\text{ N}

The two calculated forces agree closely. This numerical agreement provides direct evidence that gravitational attraction supplies the exact centripetal force needed to maintain the Moon's orbit.

Near-Earth Orbits, Geostationary Orbits, and Apparent Weightlessness

Satellites are mainly used in two kinds of orbit: near-Earth (low) orbits and geostationary orbits. You need to be able to compare them and link each orbit to its uses:

Feature	Low Earth Orbit (LEO)	Geostationary Orbit (GEO)
Typical Altitude	$200\text{ km}$ to $2,000\text{ km}$ above surface	About $36,000\text{ km}$ above the equator (orbital radius $\approx 4.2 \times 10^7\text{ m}$ )
Orbital Period	Roughly $90$ to $130$ minutes (about $92$ mins for ISS at $400\text{ km}$ )	$24$ hours ( $86,400\text{ s}$ )
Orbital Plane	Variable inclinations; frequently polar	Equatorial plane only
Direction of Orbit	Varies depending on mission	West to east (matching Earth's rotation)
Apparent Motion	Crosses rapidly across the sky	Appears stationary above a fixed spot on the equator
Primary Uses	Earth observation, weather mapping, International Space Station	Telecommunications, satellite television, continental weather tracking
Practical Advantage	High resolution images and low signal latency	Receiving dishes on homes remain permanently pointed at a fixed spot in the sky

Earth has an illustrative inclined low orbit and a larger equatorial geostationary orbit. Planetary radius, altitude and centre-to-satellite distance are marked separately.

The Three Essential Conditions for a Geostationary Orbit

For a satellite to remain in a geostationary orbit, it must satisfy three criteria:

Its orbital period must be $24$ hours ( $86,400\text{ s}$ ), the same as Earth's rotation.
It must orbit in the equatorial plane.
It must travel from west to east, rotating in the same direction as Earth.

Free Fall and Apparent Weightlessness

Astronauts floating inside an orbiting spacecraft are not beyond the reach of gravity. As shown above, gravity at $400\text{ km}$ is roughly $8.7\text{ m s}^{-2}$ .

The sensation of apparent weightlessness occurs because both the astronaut and the spacecraft are in continuous free fall towards the centre of the Earth with the same acceleration ( $a = g$ ). Because the spacecraft travels forward at high tangential speed (around $7.7\text{ km s}^{-1}$ ), Earth's surface curves away beneath it at the exact rate it falls. With the cabin floor falling at the identical rate as the astronaut's feet, there is no normal reaction force between the astronaut and the floor ( $R = 0$ ). Without a reaction force supporting the body, the astronaut experiences apparent weightlessness.

An astronaut floats clear of the spacecraft floor while both astronaut and spacecraft accelerate towards Earth at the same rate.