"Gravitation"

About

Gravitation is the fundamental force of attraction between any two masses in the universe. This chapter introduces Newton's universal law of gravitation, the concept of gravitational field, acceleration due to gravity and how it varies, Kepler's laws of planetary motion, and the motion of satellites — both natural (the Moon) and artificial.

Key Concepts

5.1 Newton's Law of Gravitation

Every particle in the universe attracts every other particle with a force directly proportional to the product of their masses and inversely proportional to the square of the distance between them.

$F=G\frac{m_1{m}_2}{r^2}$

Where $G=6.67\times 10^{-11}{\text{ N⋅m}}^2{\text{/kg}}^2$ is the universal gravitational constant.

Properties of gravitational force:

• Always attractive
• Obeys Newton's third law: $F_{12}=F_{21}$
• Acts along the line joining the centres of masses
• Independent of the medium between masses
• Central force and conservative force

Dimensional formula of G: From $F=G\frac{m_1{m}_2}{r^2}$: $[G]=\frac{[F][r^2]}{[m{]}^2}=\frac{[ML{T}^{-2}][L^2]}{[M{]}^2}=[M^{-1}{L}^3{T}^{-2}]$

Definition of G: $G$ is numerically equal to the force of attraction between two masses of 1 kg each placed 1 m apart.

5.2 Effect of Changing Masses and Distance

5.3 Acceleration Due to Gravity (g)

The acceleration experienced by a body due to Earth's gravity at the surface:

$g=\frac{GM}{R^2}$

Where $M$ = mass of Earth, $R$ = radius of Earth.

Standard value: $g\approx 9.8{\text{ m/s}}^2$

g is independent of the mass of the falling body (Galileo's observation).

Direction of g: Always vertically downward toward Earth's centre, regardless of whether the body is going up, at the top, coming down, or at rest.

5.4 Variation of g

(a) With altitude (height h above surface): $g^{\prime }=g\left(1-\frac{2h}{R}\right)\text{(for }h\ll R\text{)}$

(b) With depth (d below surface): $g^{\prime }=g\left(1-\frac{d}{R}\right)$

At the centre of Earth ($d=R$): $g^{\prime }=0$

(c) With latitude (due to Earth's rotation):

• g is maximum at poles ($R$ is smallest)
• g is minimum at equator ($R$ is largest + centrifugal effect)

$g_{\text{poles}}>g_{\text{equator}}$

(d) With shape of Earth: Earth is an oblate spheroid — $R_{\text{equator}}>R_{\text{poles}}$, so $g_{\text{poles}}>g_{\text{equator}}$ by about 0.2%.

5.5 Kepler's Laws of Planetary Motion

First Law (Law of Orbits): Each planet moves around the Sun in an elliptical orbit with the Sun at one focus.

Second Law (Law of Areas): The radius vector from the Sun to a planet sweeps out equal areas in equal intervals of time. This implies that planets move faster when closer to the Sun.

Third Law (Law of Periods): The square of the orbital period ($T$) of a planet is proportional to the cube of the semi-major axis ($a$) of its orbit: $T^2\propto a^3$

5.6 Satellite Motion

Orbital velocity: $v_o=\sqrt{\frac{GM}{r}}=\sqrt{gR}$

For a satellite close to Earth's surface: $v_o\approx 7.9\text{ km/s}$ (first cosmic velocity)

Time period: $T=2\pi \sqrt{\frac{r^3}{GM}}$

Geostationary satellite:

• Orbital period = 24 hours (same as Earth's rotation)
• Orbits in equatorial plane
• Height ≈ 36,000 km above Earth's surface

INTEXT QUESTIONS 5.1

Q1. The period of revolution of the moon around the earth is 27.3 days. The radius of moon's orbit is 3.84 × 10⁸ m (60 times the earth's radius). Calculate the centripetal acceleration of the moon and show that it is very close to the value given by 9.8 m s⁻² divided by 3600, to take account of the variation of gravity as 1/r².

Ans:

• $T=27.3\text{ days}=27.3\times 24\times 3600=2.358\times 10^6\text{ s}$
• $R=3.84\times 10^8\text{ m}$

$a=\frac{4{\pi }^{2}R}{T^2}=\frac{4{\pi }^2(3.84\times 10^8)}{(2.358\times 10^6{)}^2}$

$=\frac{3.80\times 10^{10}}{5.56\times 10^{12}}=6.8\times 10^{-3}{\text{ m/s}}^2$

Now, $\frac{g}{3600}=\frac{9.8}{3600}=2.72\times 10^{-3}{\text{ m/s}}^2$

The values are close, confirming the inverse-square law. (The small discrepancy accounts for the fact that the Moon's orbit is at 60 Earth radii, so $g_{\text{moon}}=\frac{g}{60^2}=\frac{9.8}{3600}$.)

Q2. From Eqn. (5.1), deduce dimensions of G.

Ans: From $F=G\frac{m_1{m}_2}{r^2}$:

$G=\frac{F\cdot r^2}{m_1{m}_2}$

$[G]=\frac{[ML{T}^{-2}][L^2]}{[M][M]}=[M^{-1}{L}^3{T}^{-2}]$

Q3. Using Eqn. (5.1), show that G may be defined as the magnitude of force between two masses of 1 kg each separated by a distance of 1 m.

Ans: Given $m_1=1$ kg, $m_2=1$ kg, $r=1$ m:

$F=G\frac{1\times 1}{1^2}=G$

Thus the numerical value of $G$ equals the force between two 1 kg masses placed 1 m apart.

Q4. The magnitude of force between two masses placed at a certain distance is F. What happens to F if (i) the distance is doubled without any change in masses, (ii) the distance remains the same but each mass is doubled, (iii) the distance is doubled and each mass is also doubled?

Ans: Since $F\propto \frac{m_1{m}_2}{r^2}$:

(i) $r\to 2r$ → $F^{\prime }=\frac{F}{4}$

(ii) $m_1\to 2m_1$, $m_2\to 2m_2$ → $F^{\prime }=4F$

(iii) $m_1,m_2$ doubled AND $r$ doubled → $F^{\prime }=\frac{4F}{4}=F$ (no change)

Q5. Two bodies having masses 50 kg and 60 kg are separated by a distance of 1 m. Calculate the gravitational force between them.

Ans:

• $m_1=50$ kg, $m_2=60$ kg, $r=1$ m, $G=6.67\times 10^{-11}$ N⋅m²/kg²

$F=G\frac{m_1{m}_2}{r^2}=\frac{6.67\times 10^{-11}\times 50\times 60}{1^2}$

$=6.67\times 10^{-11}\times 3000=2.0\times 10^{-7}\text{ N}$

INTEXT QUESTIONS 5.2

Q1. The mass of the earth is 5.97 × 10²⁴ kg and its mean radius is 6.371 × 10⁶ m. Calculate the value of g at the surface of the earth.

Ans:

• $M=5.97\times 10^{24}$ kg, $R=6.371\times 10^6$ m, $G=6.67\times 10^{-11}$ N⋅m²/kg²

$g=\frac{GM}{R^2}=\frac{6.67\times 10^{-11}\times 5.97\times 10^{24}}{(6.371\times 10^6{)}^2}$

$g=9.81{\text{ m/s}}^2$

Q2. Careful measurements show that the radius of the earth at the equator is 6378 km while at the poles it is 6357 km. Compare values of g at the poles and at the equator.

Ans:

• $R_e=6.378\times 10^6$ m, $R_p=6.357\times 10^6$ m

Since $g\propto \frac{1}{R^2}$:

$\frac{g_e}{g_p}={\left(\frac{R_p}{R_e}\right)}^2={\left(\frac{6.357}{6.378}\right)}^2=0.9979$

$⟹g_p\approx 1.002\times g_e$

So $g_p$ is about 0.2% larger than $g_e$.

Q3. A particle is thrown up. What is the direction of g when (i) the particle is going up, (ii) when it is at the top of its journey, (iii) when it is coming down, and (iv) when it has come back to the ground?

Ans: In all four cases, the acceleration due to gravity $g$ is vertically downward toward the Earth's centre. The direction of g never changes — it always points toward the centre of the Earth, regardless of the body's motion.

Q4. The mass of the moon is 7.3 × 10²² kg and its radius is 1.74 × 10⁶ m. Calculate the gravitational acceleration at its surface.

Ans:

• $M=7.3\times 10^{22}$ kg, $R=1.74\times 10^6$ m

$g_{\text{moon}}=\frac{GM}{R^2}=\frac{6.67\times 10^{-11}\times 7.3\times 10^{22}}{(1.74\times 10^6{)}^2}$

$g_{\text{moon}}=1.62{\text{ m/s}}^2$

This is about $\frac{1}{6}$ of Earth's g.

Terminal Exercise

1. State Newton's universal law of gravitation. Express it in vector form.

2. Define gravitational constant G. How did Cavendish determine its value?

3. Derive the expression for acceleration due to gravity at the surface of the earth: $g=\frac{GM}{R^2}$.

4. Discuss the variation of g with: (a) altitude, (b) depth, (c) latitude. Derive the relevant expressions.

5. At what height above the Earth's surface will the value of g be reduced to (a) half, (b) one-fourth of its surface value? (R = 6400 km)

6. What is the effect of Earth's rotation on the value of g? At which place on Earth would g be maximum and minimum?

7. State and explain Kepler's three laws of planetary motion.

8. Derive the expression for orbital velocity of a satellite: $v_o=\sqrt{\frac{GM}{r}}$.

9. A satellite revolves around a planet in a circular orbit. What is the work done by the gravitational force on the satellite in one complete revolution? Justify your answer.

10. Distinguish between geostationary and polar satellites. State two uses of each.

11. The mass of Jupiter is $1.9\times 10^{27}$ kg and its radius is $7.0\times 10^7$ m. Calculate the acceleration due to gravity on Jupiter's surface. Compare it with g on Earth.

12. Two bodies of masses 100 kg and 10,000 kg are placed 1 m apart. At which point on the line joining them will the gravitational field intensity be zero?

Worked Examples

Example 1: Force Between Two Masses

Problem: Calculate the gravitational force between two iron balls each of mass 1 kg placed 10 cm apart.

Solution: $F=G\frac{m_1{m}_2}{r^2}=\frac{6.67\times 10^{-11}\times 1\times 1}{(0.1{)}^2}=6.67\times 10^{-9}\text{ N}$

This is extremely small — why we don't feel gravitational attraction between everyday objects.

Example 2: g at a Height

Problem: Find the value of g at a height of 100 km above Earth's surface. (R = 6400 km, g = 9.8 m/s²)

Solution: Using $g^{\prime }=g\left(1-\frac{2h}{R}\right)$ (since $h\ll R$): $g^{\prime }=9.8\left(1-\frac{2\times 100}{6400}\right)=9.8\left(1-\frac{200}{6400}\right)$ $=9.8\left(1-0.03125\right)=9.8\times 0.96875=9.49{\text{ m/s}}^2$

Example 3: Orbital Period

Problem: A satellite orbits Earth at a height of 400 km. Find its orbital period. (R = 6400 km, g = 10 m/s²)

Solution:

• $r=R+h=6400+400=6800\text{ km}=6.8\times 10^6\text{ m}$
• $v_o=\sqrt{gR}\times \sqrt{\frac{R}{r}}$

Or using $T=2\pi \sqrt{\frac{r^3}{GM}}$ with $GM=g{R}^2$: $T=2\pi \sqrt{\frac{r^3}{g{R}^2}}=2\pi \sqrt{\frac{(6.8\times 10^6{)}^3}{10\times (6.4\times 10^6{)}^2}}\approx 5580\text{ s}\approx 93\text{ min}$

Common Mistakes

1. Thinking g changes direction during projectile motion: g ALWAYS points downward — never up, even when the body is moving up.
2. Forgetting that gravitational force is independent of the medium: Unlike electrostatic force, gravity doesn't depend on what's between the masses.
3. Confusing G and g: G is universal constant ($6.67\times 10^{-11}$); g is acceleration due to gravity (~9.8 m/s² on Earth's surface).
4. Applying $g^{\prime }=g(1-2h/R)$ for large heights: This approximation works only when $h\ll R$.
5. Thinking mass affects g: g at a location is the same for all bodies regardless of their mass.

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