**Planetary Motion** ( Class 11 Physics Gravitation )

- Ptolemy was the first scientist who studied the planetary motion.
- He gave geocentric model. It means all the planets, stars and sun revolve around the earth and earth is at the centre.
- Heliocentric model was proposed by some Indian astronomers.
- According to which all planets revolve around the sun.
- Nicholas proposed the Nicholas Copernicus modelaccording to which all planets move in circles around the sun.
- After Nicholas one more scientist named Tycho Brahe did lot of observations on planets.
- Finally came Johannes Kepler who used Tycho Brahe observations and he gave Kepler’s 3 laws of Gravitation.
- These 3 laws became the basis of Newton’s Universal law of Gravitation.

__Kepler’s 1 ^{st} law Vs. Copernicus Model__

- According to Copernicus planets move in circular motion whereas according to Kepler planets revolve in elliptical orbit around the sun.
- Copernicus model is based on one special case because circle is a special case of ellipse whereas Kepler’s laws aremore of ageneral form.
- Kepler’s law also tells us about the orbits which planets follow.

__To Show ellipse is a special form of Circle__

- Select two points F1 and F2.
- Take a pieceof string and fix its ends at F
_{1}and F_{2}. - Stretch the string taut with the help of a pencil and then draw a curve by moving the pencil keeping the string taut throughout. Fig. (a).
- The resulting closed curve is an ellipse. For any point T on the ellipse, the sum of distances from F
_{1}and F_{2}is a constant. F_{1},F_{2}are called the foci. - Join the points F
_{1}and F_{2,}and extend the line to intersect the ellipse at points P and A as shown in Fig. (a). - The centre point of the line PA is the centre of the ellipse O and the length PO = AO, which is also known as the semimajor axis of the ellipse.
- For a circle, the two foci merge onto one and the semi-major axis becomes the radius of the circle.

**Kepler’s 2 ^{nd} law: Law of Areas**

Statement:-The line that joins a planet to the sun sweeps out equal areas in equal intervals of time.

- Area covered by the planet while revolving around the sun will be equal in equal intervals of time. This means the rate of change of area with time is constant.
- Suppose position and momentum of planet is denoted by ‘r’ and ‘p’ and the time taken will be Δt.
- ΔA=1/2xrxvΔt (where vΔt is distance travelled by a planet in Δt time.)
- ΔA / Δt =1/2(
**r**x**v**) - where
- (Linear momentum) p=mv or we can write as
**v**=**p**/m- =1/2m(
**r**x**p**) - =1/2 L/2m where L= angular momentum(It is constant for any central force)

**ΔA / Δt = constant (**This means equal areas are covered in equal intervals of time).

**Kepler’s 3 ^{rd}Law: Law of periods**

Statement: –

- According to this law the square of time period of a planet is ∝ to the cube of the semi-major axisof its orbit.
- Suppose earth is revolving around the sun then the square of the time period (time taken to complete one revolution around sun) is ∝ to the cube of the semi major axis.
- It is known as Law of Periods as it is dependent on the time period of planets.
- Derivation of 3
^{rd}Law: assumption: The path of the planet is circular. - Let m=mass of planet
- M= mass of sun
- According to Newton’s Law of Gravitation:
- F= GMm/r
^{2} - F
_{c}=mv^{2}/r - where
- F
_{c}=centripetal force which helps the planet to move around sun in elliptical order. - F = Fc
- GMm/r
^{2}= mv^{2}/r where r=radius of the circle - GM/r= v
^{2}(1) - v= 2 πr/T
- Squaring both the sides the above equation
- v
^{2}=4 π^{2}r^{2}/T^{2} - putting the value (1)
- GM/r=4 πr
^{2}/T^{2} - T
^{2}= (4 π^{2}r^{3}/GM) where (4 π^{2}/GM) = constant **T**(In ellipse semi-major axis is same as radius of the circle)^{2}=r^{3}

** Problem**: – Suppose there existed a planet that went around the sun twice as fast as the earth.What would be its orbital size as compared to that of the earth?

** Answer**: Lesser by a factor of 0.63

Time taken by the Earth to complete one revolution around the Sun,

T_{e }= 1 year

Orbital radius of the Earth in its orbit, R_{e} = 1 AU

Time taken by the planet to complete one revolution around the Sun,

T_{p}=1/2T_{e}=1/2 year

Orbital radius of the planet = R_{p}

From Kepler’s third law of planetary motion, we can write:

(R_{p} /R_{e})^{3} = (T_{p}/T_{e})^{2}

R_{p} /R_{e }= (T_{p}/T_{e})^{ 2/3}

=((1/2)/1)^{ 2/3} =(0.5)^{ 2/3} = 0.63

Hence, the orbital radius of the planet will be 0.63 times smaller than that of the Earth.

** Problem**: A Saturn year is 29.5 times the earth year. How far is the Saturn from the sun, if theearth is 1.50 ×10

^{8}km away from the sun?

** Answer: **Distance of the Earth from the Sun, R

_{e}= 1.5 × 10

^{8}km = 1.5 × 10

^{11}m

Time period of the Earth = T_{e}

Time period of Saturn, T_{s} = 29. 5 T_{e}

Distance of Saturn from the Sun = R_{s}

From Kepler’s third law of planetary motion, we have

T^{2}= (4 π^{2} r^{3}/GM) ^{1/2}

For Saturn and Sun, we can write

(R_{s}^{3}/ R_{e}^{3}) = T_{s}^{2}/ T_{e}^{2}

R_{s} = R_{e}(T_{s}/ T_{e})^{2/3}

=1.5 × 10^{11}(29. 5)^{ 2/3}

=1.5 × 10^{11}x9.55

=14.32×10^{11}m

Hence, the distance between Saturn and the Sun is 1.432×10^{12}m.

** Problem:-**Let the speed of the planet at the perihelion P in Fig. be v

_{P}and

Sun-planet distance SP is r_{P}. Relate {r_{P}, v_{P}} to the corresponding quantities at

aphelion {r_{A}, v_{A}}. Will the planet take equal times to traverse BAC and CPB?

** Answer** The magnitude of the angular momentum at P is Lp = mp rp vp, since inspection tells us that r

_{p}and v

_{p}are mutually perpendicular. Similarly,

L_{A} = m_{p} r_{A} v_{A}. Fromangular momentum conservation

m_{p} r_{p} v_{p} = m_{p} r_{A} v_{A}

orv_{p}/v_{A}=r_{a}/r_{p}

Since r_{A}> r_{p}, v_{p}> v_{A} .

The area SBAC bounded by the ellipse andthe radius vectors SB and SC is larger than SBPCin Fig. From Kepler’s second law, equal areasare swept in equal times. Hence the planet willtake a longer time to traverse BAC than CPB.