Gravity& orbits

22 34P a

G M M

2

GMMF

r

Isaac Newton(1642-1727) developeda mathematical model of Gravity which predicted the elliptical orbits proposed by Kepler

r

2

2

2

22 3

1

1 cos

1

4

( )

ar

b

a

P aG M M

2a

2b

M

M

Semi-major axis

Semi-minor axis 2

GMMF

r

2

11 3 -1 -26.67 10 m kg s

GMMF

rG

Force of gravity

Planet and Solar masses

Eccentricity of ellipse

Orbital period P

Polar equation of ellipse

21

1 cos

ar

since Sun at left focus in this case

2

2

2

22 3

1

1 cos

1

4

( )

ar

b

a

P aG M M

Eccentricity of ellipse

Orbital period P

Polar equation of ellipse

Equal areas swept out in equal times

212 G( ) 1

dAM M a

dt

Johannes Kepler1571-1630

This is a constant

Orb

ital p

erio

d P

/yea

rs

Semi-major axis of orbit a / AU

32P a

32

Yr AU

P a

2

2 34P a

G M M

Radii of planets not to scale!

Mercury, Venus, Earth, Mars

Jupiter

Saturn

Uranus

Neptune

Johannes Kepler1571-1630

Kepler’s Third Law

Albert Einstein (1879-1955) proposed a radical new theory of gravity, General Relativity, in which both space & time (‘spacetime’) are curved by the presence of mass. This helped to explain anomalies in the Newtonian model such as the precession of the orbit Mercury and the amount that light is bent by massive objects (Gravitational lensing). Note General Relativity predicts the same planetary dynamics as Newton’s model when gravity is fairly weak. i.e. Newton’s model can be thought of as an approximation.

Escape velocityTo escape the gravity of a spherical astronomical body of mass M and radius R the total energy of the system must be positive at an infinite distance from the body. In other words, it will have some kinetic energy and will never be gravitationally attracted back towards the body.

For a mass m blasting off with velocity v, it will escape the gravitational influence of M if:

212 0

2

GMmmv

R

GMv

R

For Earth, the escape velocity is:

11 241

6

2

2 6.67 10 5.97 1011.2kms

6.38 10

escape

escape

GMv

R

v

It is interesting to work out the radius of a star of mass M such that the escape velocity exceeds that of the speed of light. Since this is not possible, the star becomes a Black Hole.

2

2

2

2

2

GMc

RGM

cRGM

Rc

This inequality defines the maximum radius of a Black Hole, which is called the Schwarzschild radius. This is the event horizon, or ‘point of no return’ from the centre of a Black Hole.

For the Sun to become a Black Hole ( M = 2 x 1030 kg, R = 6.96 x 108 m ) its radius would have to shrink to less than 2.97 km.

This is a mindblowing density of 1.8 x 1019 kgm-3 !

3

43 2

6

3 2

2

3

32

Black hole

Black hole

M

GM

c

c

G M

Two body Kepler problem summary

21

1 2

12

1 2

m

m m

m

m m

r R r

r R r

2

1 2

2

2

2

2

1ˆ

1 cos

( ) sin ˆˆ1 cos1 cos1

1

1

a

G m m

a

b a

b

a

r r

v r θ

ˆ ˆ ˆcos sin

ˆ ˆ ˆsin cos

r x y

θ x y

21 23 2 3

3 2 3

201 2

22 3

1 2

211 22

2 2 21 22

1 2

1 2

2 2 1 21 11 1 1 12 2

( )1 cos

(1 )

(1 )

( ) 1 cos

4

( )

G( ) 1

1

2

G m m

a

a dt

G m m

P aG m m

dAm m a

dt

Gm m aJ

m m

GmmE

aGm m

E m v m vr

Angular velocity

Orbital time

Period vs semi-major axis (Kepler III)

Equal areas swept out in equal times (Kepler II)

Angular momentum (which is a constant)

Total energy (which is a constant).

Note this equation is only true for boundelliptical orbits. Other possibilities (parabolae, hyperbolae) are possible, but they are not bound.

If E > 0 this means an orbit is not bound

Orbit initial condition for a two-body system

21 0

1 2

12 0

1 2

2

0

0

01 20 02

0

0, 0

1ˆ

1 cos

sin( ) ˆˆ1 cos1 cos1

t

m

m m

m

m m

a

G m m

a

R

r r

r r

r r

v r θ

INPUTS 1 2 0, , , , ,m m G a

ˆ ˆ ˆcos sin

ˆ ˆ ˆsin cos

r x y

θ x y

Assumes Pluto is in the same orbital plane as the other planets ....

Also that semi-major axis of all planets are aligned.

This is not true in general!

24 6

30 11

5.97 10 kg 6.38 10 m

1.99 10 kg 1AU=1.496 10 m

M R

M

Note the orbit ofPluto is not in the ‘plane of the ecliptic’-so it is actually possiblethat it can be closer toEarth than Neptune

Numerical orbit solver (Verlet method)

2( )( ) ( )

( )( )

i j

j ii j

ii

i

Gmmt

t t

tt

m

iFr r

Fa

i.e. constant acceleration motion between fixed time steps DtNote velocity calculation incorporates updated acceleration, determined usingthe updated position vector. This increases the accuracy of the method.

Newton’s Law of universal gravitation

Newton’s Second Law

Update position and velocity vectors

212

12

( ) ( ) ( ) ( )

( ) ( ) ( ) ( )

i i i

i i i i

t t t t t t t

t t t t t t t

ir r v a

v v a a

Mass of star (solar masses) 1Mass of planet (Earth masses) 1Semi-major axis of orbit /AU 1Orbital eccentricity 0.4Initial polar angle /degrees 180

timestep /years 0.001

period /years 1.0008977

‘Numerical artefacts' due to timestep being too large result in an erroneous orbit

This is a problemwith numeric solutions!

timestep = 0.05 years.

Using the Verlet method to ‘solve’ a three-body problemi.e. the motion of a planet within the mutual orbitsof a binary star system

This time the planet starts outside both the orbits of thebinary star system

Orbits don’t have to be bound .... This one is unstable. Aftera few orbits, the planet is ejected from the system!

Complicated orbits can result if the planet is closeto both stars!