Kepler's Laws of Planetary Motion

Table of Contents
1. Kepler's Laws of Planetary Motion
2. Connection with Newton's Law of Gravitation
3. Newton's Third Law of Motion and Gravitation
4. Free Fall
5. Acceleration due to Gravity (g)
6. Variation in the Value of g (High Order Thinking)
7. Do You Know?
8. Difference between "g" and "G"
9. Motion of Objects under the Influence of Earth's Gravity
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Kepler's Laws of Planetary Motion

Johannes Kepler (1571-1630) was an astronomer who discovered three empirical laws that describe the motion of planets around the Sun. These laws apply to any object orbiting a much more massive central body and so also describe motion of natural and artificial satellites. The three laws are stated and explained below, followed by the connection with Newton's law of gravitation and related ideas useful for class 9 students.

Kepler's First Law

Statement: The orbit of a planet around the Sun is an ellipse with the Sun at one of its two foci.

Explanation and terms:

• Ellipse: A closed curve for which the sum of distances from two fixed points (the foci) to any point on the curve is constant.
• Semi-major axis (a): Half of the longest diameter of the ellipse.
• Eccentricity (e): A number between 0 and 1 that measures how much the ellipse differs from a circle (e = 0 is a circle).
• For planetary orbits the eccentricity is usually small, so most planetary orbits are nearly circular but still elliptical.

Kepler's Second Law

Statement: A line joining a planet and the Sun sweeps out equal areas in equal intervals of time.

Explanation:

• This means that a planet moves faster when it is nearer to the Sun (at perihelion) and slower when it is farther from the Sun (at aphelion), so that the area swept in a given time is constant.
• The law is equivalent to the conservation of angular momentum for a central force: the torque about the central body is zero, so the areal velocity is constant.

Kepler's Third Law

Statement: The square of the orbital period (T) of a planet is directly proportional to the cube of the semi-major axis (a) of its orbit.

Mathematical form:

• a³ ∝ T²
• or a³ = constant × T² (the constant is the same for all planets orbiting the same central body, e.g. the Sun)

Interpretation:

• If we measure a in metres and T in seconds for planets around the Sun, the constant equals 4π² / (GM⊙), where M⊙ is the mass of the Sun and G the universal gravitational constant. Thus Kepler's third law links orbital motion to the central mass.

 
 where r = Mean distance of planet from the sun
 and T = Time period of the planet (around the sun)
 Through Kepler gave the laws of planetary motion but he could not give a theory to explain the motion of planets. It was Newton who showed that the cause of the motion of planets is the gravitational force which the sun exerts on them. In fact, Newton used the Kepler's third law of planetary motion to develop the law of universal gravitation.

Connection with Newton's Law of Gravitation

Newton's Inverse-square Law of Gravitation

Statement: Every pair of point masses attracts each other with a force whose magnitude is directly proportional to the product of the masses and inversely proportional to the square of the distance between them.

Mathematical form:

• F = G (m₁ m₂) / r², where G is the universal gravitational constant.

 
 Consider a planet of mass m moving with a velocity (or speed) v around the sun in a circular orbit of radius r, A centripetal force F acts on the orbiting planet (due to the sun) which is given by :

Using Newton's law of gravitation together with the condition for circular motion (or with appropriate centripetal relation for near-circular orbits) one can derive Kepler's third law and show why T² ∝ a³. This derivation shows that the constant in Kepler's third law is related to G and the mass of the central body.

Simple derivation (outline)

For a planet of mass m orbiting the Sun of mass M in a circular orbit of radius r with speed v:

centripetal force required = m v²/r

gravitational force = G M m / r²

Equating the two:

m v²/r = G M m / r²

Cancel m and rearrange:

v² = G M / r

Circular speed and period relation: v = 2π r / T

Substitute v:

(2π r / T)² = G M / r

So:

T² = (4π²/G M) r³

This gives T² ∝ r³ and the constant 4π²/(G M) is the same for all planets orbiting the same central mass M.

 The mass m of a given planet is constant

 

The factor 2π is a constant

Now, taking square on both sides

If we multiply as well as divide the right side of this relation by r  

The factor    is constant by Kepler's third law.    

by putting 1/r  in place of v² in relation        or   

Newton's Third Law of Motion and Gravitation

Newton's third law of motion-that every action has an equal and opposite reaction-also applies to gravitational forces. When the Earth attracts an object with a gravitational force, the object attracts the Earth with an equal and opposite force.

According to Newton's second law,
 Force = Mass × Acceleration
 F = ma
 Acceleration

Because the mass of the Earth is extremely large compared to everyday objects, the acceleration produced in the Earth by this force is negligibly small and is not noticeable.

Free Fall

 
 Any object dropped from some height always falls towards the earth. If a feather and a stone are dropped from the top of a tower, it is observed that feather falls onto the ground much later than the stone. So, it was thought that object of different masses dropped from same height take different times to reach the ground 

Galileo's experiments (for example at the Tower of Pisa) showed that in the absence of air resistance all bodies dropped from the same height fall with the same acceleration, regardless of their masses. A feather falls more slowly in air because of air resistance, but in vacuum a feather and a stone fall together.

Conclusion: If air resistance is neglected, all bodies fall with the same acceleration due to gravity.

Definition of Free Fall

A body is said to be in free fall when the only force acting on it is the gravitational force of the Earth. The acceleration of a freely falling body is the acceleration due to gravity and is denoted by g.

Experimental Verification

After Galileo, Robert Boyle used a vacuum pump to remove air from a chamber containing a feather and a lead ball. When the chamber was inverted, both fell to the bottom at the same time, confirming that in vacuum the acceleration is independent of mass.

  

Acceleration due to Gravity (g)

Definition: The acceleration with which an object falls toward the Earth under the influence of Earth's gravitational pull (neglecting air resistance) is called the acceleration due to gravity and denoted by g.

All bodies near the Earth's surface fall with (approximately) the same constant acceleration g. Its average value at Earth's surface is about 9.8 m s^-2.

To derive the value of g

Consider a body of mass m at distance R from the centre of the Earth (R is the radius of the Earth if the body is on the surface). According to Newton's law of gravitation:

F = G (M m) / R²

If this force produces an acceleration a in mass m, then:

F = m a = m g

Equate the two expressions for F:

m g = G (M m) / R²

Cancel m:

g = G M / R²

 

Using values:

M (mass of Earth) ≈ 6.0 × 10²⁴ kg

R (radius of Earth) ≈ 6.4 × 10⁶ m

G ≈ 6.67 × 10^-11 N m² kg^-2

 

Value of g on the Moon

Mass of Moon = 7.4 × 10²² kg

Radius of Moon = 1,740 km = 1.74 × 10⁶ m

Using g = G M / R² for the Moon gives a value ≈ 1.6 m s^-2, which is about 1/6 of the value of g on Earth.

Mass and Average Density of the Earth

From g = G M / R² we get:

M = g R² / G

Once M is known, the average density ρ of the Earth can be found from:

ρ = 3M / (4 π R³)

 ∴

Calculation: acceleration on the Moon (outline)

Using M_moon = 7.4 × 10²² kg, R_moon = 1.74 × 10⁶ m and G = 6.7 × 10^-11 N m² kg^-2, compute g_moon = G M_moon / R_moon². This yields g_moon ≈ 1.6 m s^-2, close to 1/6 of Earth's g.

Acceleration due to gravity on the moon,  

 

Variation in the Value of g (High Order Thinking)

1. Variation with latitude and shape of the Earth

The ideal expression g = G M / R² assumes a spherical Earth. In reality the Earth is an oblate spheroid (slightly flattened at the poles and bulged at the equator). The equatorial radius R_E is slightly greater than the polar radius R_P.

Since g ∝ 1/R², a larger radius gives a smaller g. Therefore, g is greater at the poles and smaller at the equator (ignoring the small additional effect of Earth's rotation). The combined effect of Earth's rotation further reduces the effective g at the equator.

Now from equation (1), value of 'g' at equator is given by

Value of 'g' at pole is given by

2. Variation with altitude (height above Earth's surface)

For a height h above the surface, distance from Earth's centre = R + h. The acceleration due to gravity at height h is:

g_h = G M / (R + h)²

Dividing by g at the surface:

g_h / g = (R / (R + h))²

For small h compared to R, expand approximately:

g_h ≈ g (1 - 2h / R)

This shows that g decreases with increasing height above the Earth.

3. Variation with depth below the Earth's surface

Assuming the Earth has uniform density (a simplifying assumption often used in class 9), the value of g at a depth d below the surface is due only to the mass contained within radius (R - d). The result is:

g_d = g (1 - d / R)

This shows g decreases linearly with depth (under the uniform density assumption) and at the Earth's centre (d = R) the value becomes zero.

  

 

  

Do You Know?

• The acceleration due to gravity of a planet depends on its mass and radius. It is larger when mass is large and radius is small.
• The average value of g at the Earth's surface is 9.8 m s^-2.
• g decreases with depth below the surface and with height above the surface.
• g is greater at the poles and smaller at the equator (both because of shape and because of centrifugal effect of rotation).
• The value of g is zero at the centre of the Earth (assuming spherical symmetry).
• Among the planets mentioned, acceleration due to gravity is smallest on Mercury and largest on Jupiter (because of their masses and radii).
• Acceleration due to gravity is independent of the mass, shape, or size of a falling body (neglecting air resistance): light and heavy bodies fall with the same acceleration.
• The fractional rate of decrease of g with height is twice that with depth (for small variations), i.e. for small h, Δg/g ≈ -2h/R above the surface while for small depth d, Δg/g ≈ -d/R below the surface.
• If the Earth's rotation rate increases, the effective value of g at places away from the poles decreases because the centrifugal acceleration R ω² acts outward; at the poles centrifugal effect is zero.
• If the Earth stopped rotating, the value of g at the equator would increase by an amount equal to R ω² (numerically about 0.034 m s^-2), with no change at the poles due to rotation.

3. The value of g decreases with height.   

Difference between "g" and "G"

Motion of Objects under the Influence of Earth's Gravity

When bodies fall under gravity, they experience acceleration g (≈ 9.8 m s^-2) downward. When bodies are thrown upward against gravity, they decelerate at rate g. The standard equations of motion apply with acceleration a replaced by ±g for vertical motion.

For motion under constant acceleration use the following relations with a = ±g as appropriate:

• v = u + a t → for free fall set a = g (downwards)
• v = u - g t → for upward motion where acceleration opposes motion
• s = u t + ½ a t² → replace a by g and s by h when vertical
• v² - u² = 2 a s → for vertical motion replace a by ±g and s by h

Thus, for a body projected vertically upward with initial speed u:

Maximum height h is given by v² - u² = -2 g h, where v = 0 at maximum height, hence h = u² / (2 g).

This chapter has presented Kepler's three laws, explained their meaning, shown how Newton's law of gravitation connects with Kepler's third law, and reviewed the basic ideas of free fall and acceleration due to gravity including their variations and numerical values relevant for class 9 students.