D1-1. Newton's Law of Gravitation and Gravitational Field Strength

1. Newton's Law of Gravitation

The gravitational force between two bodies outside a uniform field, e.g. between the Earth and the Sun, is defined by Newton's Law of Gravitation.

  • Newton's Law of Gravitation states that: The gravitational force between two point masses is proportional to the product of the masses and inversely proportional to the square of their separation.
  • Occurs between any two point masses.
  • In equation form, this can be written as:
    $$F = G\frac{Mm}{r^2}$$
  • Gravity is always attractive. There are no negative masses.
M F' F = F' m F r

2. Gravitational Field Strength & Field Lines

A gravitational field is defined as a region of space where a test mass experiences a force due to the gravitational attraction of another mass. The direction of the gravitational field is always towards the centre of the mass causing the field, and gravitational forces are always attractive.

  • The gravitational field strength ($g$) at a point is defined as the force per unit mass experienced by a small test mass at that point.
  • It is a vector pointing towards $M$.
  • By equating $F = mg$ with Newton's law of gravitation, we get:
    $$g = \frac{F}{m} = G\frac{M}{r^2}$$

Notation Key & Units:

  • $F$ = gravitational force ($\text{N}$)
  • $G$ = gravitational constant ($6.67 \times 10^{-11}\text{ N m}^2\text{ kg}^{-2}$)
  • $M, m$ = masses ($\text{kg}$)
  • $r$ = distance between centers ($\text{m}$)
  • $g$ = field strength ($\text{N kg}^{-1}$ or $\text{m s}^{-2}$)

Non-Uniform Field of an Isolated Mass

POINT MASS FIELD LINES ARE RADIALLY INWARDS

Uniform Field Where g is the Same

EARTH'S SURFACE FIELD LINES EQUALLY SPACED AND PARALLEL

3. Superposition of Gravitational Fields

Assume two masses $M_1 > M_2$:

  • Because gravity is always attractive, the fields oppose each other in the space between the masses.
  • There is a Neutral Point ($\vec{g}=0$) where their gravitational pulls perfectly cancel out. It's on the line joining $M_1$ and $M_2$.
× Neutral Point g=0 M₁ M₂ Longer Shorter

4. Orbital Motion, Kepler's Laws & Escape Speed

Kepler's First & Second Laws

  • First Law: The orbit of a planet is an ellipse, with the Sun at one of the two foci.
  • Second Law: A line segment joining the Sun to a planet sweeps out equal areas in equal time intervals. A consequence is that planets move faster nearer the Sun and slower further away.
Area 1 Area 2 Sun Time t Time t If Area 1 > Area 2, the times are NOT equal.

Orbital Motion & Kepler's Third Law

  • For a circular orbit, the gravitational force provides the exact centripetal force required to keep it in orbit.
$$F_G = F_c \implies \frac{GMm}{r^2} = \frac{mv^2}{r}$$
  • Orbital Velocity: $v = \sqrt{\frac{GM}{r}}$
  • Kepler's Third Law: By substituting $v = \frac{2\pi r}{T}$, we get:
    $$T^2 = \left(\frac{4\pi^2}{GM}\right)r^3$$
    This proves $T^2 \propto r^3$.
M m r v FG

5. Examples

Example 1: Mass of the Earth

Problem: A satellite of mass $6500$ kg orbits at $2000$ km above the Earth's surface. The gravitational force between the Earth and the satellite is $37\text{ kN}$. Calculate the mass of the Earth. (Radius of the Earth = $6400\text{ km}$).

Solution:

Step 1: Let $m_1$ be the satellite's mass and $m_2$ be the Earth's mass. Use Newton's Law of Gravitation and rearrange to get $$ F_G = \frac{Gm_1m_2}{r^2} \implies m_2 = \frac{r^2 F_G}{G m_1} $$

Step 2: Calculate the distance $r$ from the centre of the Earth to the satellite $$ r = 2000\text{ km} + 6400\text{ km} = 8400\text{ km} = 8400 \times 10^3\text{ m} $$

Step 3: Substitute the values:
$$ m_2 = \frac{(8400 \times 10^3)^2 \times 37 \times 10^3}{6.67 \times 10^{-11} \times 6500} \approx \mathbf{6.0 \times 10^{24}\text{ kg}} $$

EXAMINER TIP

A common mistake in exams is to forget to add together the distance from the surface of the planet and its radius to obtain the value of $r$. The distance $r$ is always measured from the centre of the mass.

Example 2

Problem: A planet has half the mass and half the radius of the Earth. What is the gravitational field strength at the surface of the planet? Suppose the gravitational field strength at the surface of the Earth is $10\text{ N kg}^{-1}$.

Solution:

The equation for gravitational field strength is $$ g = \frac{GM}{R^2} $$ If mass is halved, $g$ scales by $1/2$. If radius is halved, $g$ scales by $1/(1/2)^2 = 1/(1/4) = 4$. Total scaling factor: $\dfrac{1}{2} \times 4 = 2$. Therefore, the field strength on the new planet is $2 \times 10 = \mathbf{20\text{ N kg}^{-1}}$.

Example 3

Problem: Planet X has mass $M$ and radius $R$. Planet Y has mass $2M$ and radius $3R$. The gravitational field strength at the surface of planet X is $g$. What is the gravitational field strength at the surface of planet Y in terms of $g$?

Solution:

For planet X: $$ g = \frac{GM}{R^2} $$ For planet Y: $$ g_Y = \frac{G(2M)}{(3R)^2} = \frac{2GM}{9R^2} $$ Factoring out the constants for planet X gives $$ g_Y = \frac{2}{9} \left(\frac{GM}{R^2}\right) = \mathbf{\frac{2}{9}g} $$

Example 4

Problem: What is the correct definition of gravitational field strength?
A. The mass per unit weight
B. The weight of a small test mass
C. The force acting on a small test mass
D. The force per unit mass acting on a small test mass

Solution:

Gravitational field strength ($g$) is defined by the equation $g = \frac{F}{m}$, which translates into words as "force ($F$) per unit mass ($m$)". The test mass must be small so it does not distort the field being measured. The correct answer is D.

Example 5

Problem: A satellite X of mass $m$ orbits the Earth with a period $T$. What will be the orbital period of satellite Y of mass $2m$ occupying the same orbit as X?

Solution:

According to Kepler's Third Law $$ T^2 = \frac{4\pi^2}{GM_{earth}} r^3 $$ the orbital period of a satellite depends ONLY on the radius of the orbit $r$ and the mass of the central body being orbited $M_{earth}$. The mass of the satellite itself $m$ perfectly cancels out in the derivation $$ \left(\frac{GM_{earth}m}{r^2} = \frac{mv^2}{r}\right) $$ Since they occupy the same orbit, the period is identical: $T$.

Example 6

Problem:
(a) (i) Define gravitational field strength. (ii) State the SI unit.
(b) A planet orbits the Sun in a circular orbit with orbital period $T$ and orbital radius $R$. The mass of the Sun is $M$.
(i) Show that $T = \sqrt{\dfrac{4\pi^2 R^3}{GM}}$.
(ii) The Earth’s orbit around the Sun is almost circular with radius $1.5 \times 10^{11}$ m. Estimate the mass of the Sun.

Solution:

(a) (i): Gravitational field strength is the gravitational force exerted per unit mass on a small point mass.
(a) (ii): $\text{N kg}^{-1}$ or $\text{m s}^{-2}$.

(b) (i): Equate gravitational force to centripetal force $$ \frac{GmM}{R^2} = \frac{mv^2}{R} $$ Substitute orbital velocity $v = \dfrac{2\pi R}{T}$: $$ \frac{GM}{R^2} = \frac{\left(\frac{2\pi R}{T}\right)^2}{R} = \frac{4\pi^2 R}{T^2} \implies T = \sqrt{\frac{4\pi^2 R^3}{GM}} $$

(b) (ii): The period of the Earth's orbit is 1 year. Convert to seconds:
$$ T = 365.25 \times 24 \times 60 \times 60 \approx 3.16 \times 10^7\text{ s} $$ Rearrange the equation from (b)(i) to solve for $M$: $$ M = \frac{4\pi^2 R^3}{G T^2} $$ Substitute the values: $$ M = \frac{4\pi^2 (1.5 \times 10^{11})^3}{(6.67 \times 10^{-11}) (3.16 \times 10^7)^2} \approx \mathbf{2.0 \times 10^{30}\text{ kg}} $$