# For a planet moving around sun in elliptical path

with semi major axis a and semi minor axis b find the time period by

deriving keplers law first

1
deriving keplers law first

by avinash14 31.10.2014

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with semi major axis a and semi minor axis b find the time period by

deriving keplers law first

1
deriving keplers law first

by avinash14 31.10.2014

Log in to add a comment

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Keplers 2nd law says that a planet revolving around the Sun at the focus of the elliptical orbit, sweeps the same areas in the same period of time. We want to prove that.

The area ΔA swept (in time Δt) by a planet of mass m moving with instantaneous velocity v at a instantaneous distance r from the Sun,

ΔA = 1/2 (r Χ v Δt ) as the Δs = arc length = v Δt

the quantities in bold are vectors.

ΔA/Δt = 1/2 (r Χ v) = 1/2 (r Χ p/m) = L / 2m

L is the angular momentum vector of the planet.

As there is not external force on the planet and Sun system, the total angular momentum is a constant. Hence L is constant. That means ΔA/Δt is constant. Hence Area is directly proportional to time or Area swept is constant for a constant duration of time.

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Time period T:

Area swept = ( L/2m ) t

Area covered / swept by the planet in one year T of planet =

A = area of ellipse = π a b = (L/2m) T

hence*T = (2 π a b m) / L* --- equation 1

Expression for the angular momentum L can be obtained by expressing the equation of motion for the planet and solving the 2nd order differential equation. We use the polar coordinates (r,Ф) to express the elliptical orbit.

d²r/dt² - r ω² = - GM/r² , substitute L = m r² ω

d²r/dt² = L²/(m² r³) - GM/r²

Elliptical orbit is given by r (1+e cosФ) = a (1-e²) = b²/a

Solving the differential equation, we get

L² = GMm² a (1-e²) and b² = a² (1 - e²)

From equation 1,

T² = 4π²a²b²m² / L² = 4π²a²b²m² / [ GMm² a (1-e²) ]

** Hence, T² = 4π² a³ /(GM)**

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Simplified solution assuming that planet moves in a circular orbit, with radius = semi major axis a.

Here we assume that semi major axis and semi minor axis do not differ a lot and nearly same. We take r = a = semi major axis and assume the planet going in a circular orbit. We take the Sun as the origin and the inertial frame of reference.

Centripetal acceleration of the planet exerted by the Sun on the planet =

= v² / r = r ω² = r (2π/T)² = 4π² r /T² , where T is the time period.

= gravity = G M_s / r² , where M_s = mass of Sun

Hence, G M_s / r² = 4 π² r / T²

T² = 4 π² r³ / G M_s ,

This is the same formula for the time period of revolution of the planet around the Sun at the focus of Elliptical orbit.

The area ΔA swept (in time Δt) by a planet of mass m moving with instantaneous velocity v at a instantaneous distance r from the Sun,

ΔA = 1/2 (r Χ v Δt ) as the Δs = arc length = v Δt

the quantities in bold are vectors.

ΔA/Δt = 1/2 (r Χ v) = 1/2 (r Χ p/m) = L / 2m

L is the angular momentum vector of the planet.

As there is not external force on the planet and Sun system, the total angular momentum is a constant. Hence L is constant. That means ΔA/Δt is constant. Hence Area is directly proportional to time or Area swept is constant for a constant duration of time.

===================================

Time period T:

Area swept = ( L/2m ) t

Area covered / swept by the planet in one year T of planet =

A = area of ellipse = π a b = (L/2m) T

hence

Expression for the angular momentum L can be obtained by expressing the equation of motion for the planet and solving the 2nd order differential equation. We use the polar coordinates (r,Ф) to express the elliptical orbit.

d²r/dt² - r ω² = - GM/r² , substitute L = m r² ω

d²r/dt² = L²/(m² r³) - GM/r²

Elliptical orbit is given by r (1+e cosФ) = a (1-e²) = b²/a

Solving the differential equation, we get

L² = GMm² a (1-e²) and b² = a² (1 - e²)

From equation 1,

T² = 4π²a²b²m² / L² = 4π²a²b²m² / [ GMm² a (1-e²) ]

====================================

Simplified solution assuming that planet moves in a circular orbit, with radius = semi major axis a.

Here we assume that semi major axis and semi minor axis do not differ a lot and nearly same. We take r = a = semi major axis and assume the planet going in a circular orbit. We take the Sun as the origin and the inertial frame of reference.

Centripetal acceleration of the planet exerted by the Sun on the planet =

= v² / r = r ω² = r (2π/T)² = 4π² r /T² , where T is the time period.

= gravity = G M_s / r² , where M_s = mass of Sun

Hence, G M_s / r² = 4 π² r / T²

T² = 4 π² r³ / G M_s ,

This is the same formula for the time period of revolution of the planet around the Sun at the focus of Elliptical orbit.