1. Force

Force is defined as:

\[ F = ma \]

where:

• \(F\) = force
• \(m\) = mass
• \(a\) = acceleration

The SI unit of force is Newton (N).

Four Fundamental Forces

1. Gravitational force
2. Electromagnetic force
3. Strong nuclear force
4. Weak nuclear force

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2. Newton's Law of Universal Gravitation

Newton's law of gravitation states:

Every two bodies in the universe attract each other with a force that is directly proportional to the product of their masses and inversely proportional to the square of the distance between them.

For two bodies of masses \(m_1\) and \(m_2\), separated by distance \(r\):

\[ F \propto m_1m_2 \]

and

\[ F \propto \frac{1}{r^2} \]

Therefore,

\[ \boxed{F = G\frac{m_1m_2}{r^2}} \]

where \(G\) is the universal gravitational constant.

Value of \(G\)

\[ \boxed{G = 6.674\times10^{-11}\ \text{N m}^2\text{kg}^{-2}} \]

\(G\) is universal, meaning its value is the same everywhere.

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3. Gravitational Constant \(G\)

From

\[ F = G\frac{m_1m_2}{r^2} \]

we get

\[ \boxed{G=\frac{Fr^2}{m_1m_2}} \]

The gravitational constant is defined as the gravitational force between two bodies of unit mass separated by unit distance.

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4. Relation Between \(g\) and \(G\)

Consider a body of mass \(m\) near the surface of Earth.

Let:

• \(M_E\) = mass of Earth
• \(R_E\) = radius of Earth
• \(m\) = mass of the body
• \(g\) = acceleration due to gravity

The force on the body due to gravity is:

\[ F_1 = mg \]

From Newton's law of gravitation:

\[ F_2 = G\frac{M_E m}{R_E^2} \]

Since both expressions represent the same gravitational force:

\[ mg = G\frac{M_E m}{R_E^2} \]

Cancel \(m\):

\[ \boxed{g = G\frac{M_E}{R_E^2}} \]

Therefore,

\[ \boxed{g=\frac{GM_E}{R_E^2}} \]

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5. Acceleration Due to Gravity

The acceleration due to gravity is approximately:

\[ \boxed{g=9.8\ \text{m/s}^2} \]

Dimensional Formula

\[ g = \frac{\text{length}}{\text{time}^2} \]

Therefore,

\[ \boxed{[g]=[LT^{-2}]} \]

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6. Weight of a Body

Weight is the gravitational force exerted by Earth on a body.

\[ \boxed{W=mg} \]

where:

• \(W\) = weight
• \(m\) = mass
• \(g\) = acceleration due to gravity

The SI unit of weight is Newton (N).

Dimensional Formula

\[ [W]=[MLT^{-2}] \]

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7. Dimensional Formulae

Gravitational Force

\[ \boxed{[F]=[MLT^{-2}]} \]

Gravitational Constant

From

\[ G=\frac{Fr^2}{m_1m_2} \]

\[ [G] = \frac{[MLT^{-2}][L^2]}{[M^2]} \]

Therefore,

\[ \boxed{[G]=[M^{-1}L^3T^{-2}]} \]

Acceleration Due to Gravity

\[ \boxed{[g]=[LT^{-2}]} \]

Weight

\[ \boxed{[W]=[MLT^{-2}]} \]

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Solved Problems

8. Gravitational Force Between Two Bodies

Question

Find the gravitational force between two bodies, each of mass \(20\,\text{kg}\), separated by \(50\,\text{cm}\).

Given

\[ m_1=m_2=20\,\text{kg} \]

\[ r=50\,\text{cm}=0.5\,\text{m} \]

\[ G=6.67\times10^{-11}\ \text{N m}^2\text{kg}^{-2} \]

Using:

\[ \begin{aligned} F &= G\frac{m_1m_2}{r^2} \\ F &= \frac{(6.67\times10^{-11})(20)(20)}{(0.5)^2} \\ F &=(6.67\times10^{-11})(1600) \end{aligned} \]

Therefore,

\[ \boxed{F=1.0672\times10^{-7}\ \text{N}} \]

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9. Acceleration Due to Gravity on the Moon

Question

What will be the acceleration due to gravity on the Moon if:

• Radius of Moon = \(\frac14\) radius of Earth
• Mass of Moon = \(\frac1{80}\) mass of Earth

Given

\[ R_M=\frac{R_E}{4} \]

\[ M_M=\frac{M_E}{80} \]

We know:

\[ g=G\frac{M}{R^2} \]

For Earth:

\[ g_E=G\frac{M_E}{R_E^2} \]

For Moon:

\[ g_M=G\frac{M_M}{R_M^2} \]

Therefore,

\[ \begin{aligned} \frac{g_M}{g_E} &= \frac{M_M}{M_E} \frac{R_E^2}{R_M^2} \end{aligned} \]

Substituting:

\[ \begin{aligned} \frac{g_M}{g_E} &= \frac1{80} \left(\frac{R_E}{R_E/4}\right)^2 \\ &= \frac1{80}(4)^2 \\ &= \frac{16}{80} \\ &= \frac15 \end{aligned} \]

Hence,

\[ g_M=\frac{g_E}{5} \]

Taking \(g_E=9.8\,\text{m/s}^2\):

\[ g_M=\frac{9.8}{5} \]

\[ \boxed{g_M=1.96\ \text{m/s}^2} \]

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10. Gravitational Force Between the Sun and Jupiter

Question

The mass of Jupiter is \(1.9\times10^{27}\,\text{kg}\), the mass of the Sun is \(1.99\times10^{30}\,\text{kg}\), and their mean distance is \(7.8\times10^{11}\,\text{m}\). Calculate the gravitational force exerted by the Sun on Jupiter.

Given

\[ m_J=1.9\times10^{27}\,\text{kg} \]

\[ m_S=1.99\times10^{30}\,\text{kg} \]

\[ r=7.8\times10^{11}\,\text{m} \]

\[ G=6.67\times10^{-11}\ \text{N m}^2\text{kg}^{-2} \]

Using:

\[ F=G\frac{m_Jm_S}{r^2} \]

\[ F= \frac{ (6.67\times10^{-11}) (1.9\times10^{27}) (1.99\times10^{30}) }{ (7.8\times10^{11})^2 } \]

Therefore,

\[ \boxed{F\approx4.15\times10^{23}\ \text{N}} \]

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11. Gravitational Force Between an Electron and a Proton

Question

Find the gravitational force between an electron and a proton if their separation is \(1\,\text{A}\).

Given

Mass of electron:

\[ m_e=9.1\times10^{-31}\,\text{kg} \]

Mass of proton:

\[ m_p=1.67\times10^{-27}\,\text{kg} \]

Separation:

\[ r=1\,\text{A} \]

Since:

\[ 1\,\text{A}=10^{-10}\,\text{m} \]

we have:

\[ r=10^{-10}\,\text{m} \]

Using:

\[ F=G\frac{m_em_p}{r^2} \]

\[ F= \frac{ (6.67\times10^{-11}) (9.1\times10^{-31}) (1.67\times10^{-27}) }{ (10^{-10})^2 } \]

Therefore,

\[ \boxed{F\approx1.014\times10^{-47}\ \text{N}} \]

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12. Important Formula Sheet

Quantity	Formula	SI Unit	Dimension
Force	\(F=ma\)	N	\([MLT^{-2}]\)
Gravitational force	\(F=G\dfrac{m_1m_2}{r^2}\)	N	\([MLT^{-2}]\)
Gravitational constant	\(G=\dfrac{Fr^2}{m_1m_2}\)	N m^2 kg^-2	\([M^{-1}L^3T^{-2}]\)
Acceleration due to gravity	\(g=\dfrac{GM}{R^2}\)	m/s^2	\([LT^{-2}]\)
Weight	\(W=mg\)	N	\([MLT^{-2}]\)

Constants

\(\text{G}\)	\(6.674 \times 10^{-11} \text{N} \text{m}^{2} \text{kg}^{-2}\)
\(\text{g}\)	\(\approx 9.8 \text{m/s}^2\)

\[ \boxed{G=6.674\times10^{-11}\ \text{N m}^2\text{kg}^{-2}} \]

\[ \boxed{g\approx9.8\ \text{m/s}^2} \]

\[ \boxed{1\,\text{A}=10^{-10}\,\text{m}} \]