Thursday, 21 April 2011

Physics : Planetary Motion

Note: This note requires basic knowledge of calculus.
Planetary Motion
In planetary motion, the gravitational force is given by F = GMm/r2. By W = Fs,
dW/dr = d(Fs)/dr = d(Fr)/dr = F,   W = ∫F dr.
The gravitational potential energy (U) = ∫F dr = ∫ (GMm/r2) dr = -GMm/r + C.
Since we take the point of infinitely far as zero, so C = 0. Therefore U = -GMm/r
When r is near to Earth’s surface, work done to raise an object by height h (h<<R)is given by:
W = U’ – U = -GMm(1/(R+h) – 1/R) = GMmh/(R2+Rh) ≈ (GMm/R2)(mh) = mgh.
Therefore P.E. = mgh is a expression when it’s near to Earth’s surface.
Negative U implies that work have to be done to increase the distance between two object (then the EPE in the system of these two objects increases, i.e., less negative)
Equating mechanical energy (K.E. + E.P.E.), we say mvA2/2 - GMm/rA = mvB2/2 - GMm/rB.
Now consider circular orbital motion around a planet: GMm/r2 = mv2/2, v = (GM/r)0.5.
The mechanical energy of the orbiting object = E = mv2/2 – GMm/r = m(GM)/2r – GMm/r = U/2.
(Note: this is similar to the case of electron orbiting the nucleus. Also, this is applicable only when this is circular orbit.)
Escape velocity
We say an object is escaped from the affection by the gravity due to a mass when the mechanical energy of the object is positive, i.e., |K.E.|>|E.P.E.|, then at infinitely far it still has positive velocity (take the direction of g-field be negative). Mathematically mv2/2 ≥ GMm/r, then we have the minimum escaping velocity vesc = (2GM/R)0.5 = 20.5vorbiting.
Considering the escaping velocity and orbiting velocity, we know that:
a)       If vproj < (GM/r)0.5, the object performs projectile motion and falls to the ground.
b)       If vproj = (GM/r)0.5, the object performs circular orbiting around the mass.
c)       If (GM/r)0.5 < vproj < (2GM/r)0.5, the object performs elliptical orbiting around the mass.
d)       If vproj > (2GM/r)0.5, the object escape from the mass and never returns.
For Earth, the escaping velocity is approximately 11.2km s-1.
Elliptical orbit
Ellipse contains two foci. The sum of distance from a point on the ellipse to the two foci is the same for all point on the ellipse. For elliptical orbiting, the mass will be at one of the focus.
The point that is the farthest from the mass is called perihelion (for objects orbiting Earth, it’s called perigee) and the closest point is aphelion (apogee). The length of semi-major axis is given by major axis/2 or the (distance from perihelion + distance from aphelion)/2.
Kepler’s law of planetary motion
First law: The orbit of a planet about a star is an ellipse with the star at one focus.
Proof: if it’s not located at one of the focus, the planetary motion does not consistent with Newton’s law of motion. However it’s omitted here due to limited knowledge. 
Second law: A line joining a planet and its star sweeps out equal areas during equal intervals of time.
Proof: Consider the angular momentum : L = r × p = m(r × v), where × is the cross product.
Since v = dr/dt, L = m(r × dr/dt), dL/dt = m(dr/dt × dr/dt) = 0 (since parallel vector has zero cross product). Therefore L is a fixed value independent of its location in the orbit.
A = r2/2, dA = r dr/2 = (r × dr/dt) dt/2 = L(dt)/2m, therefore dA/dt = L/2m which is also fixed, independent of the location of the planar in the orbit.
Third law: The square of the orbiting period of an orbiting planet is directly proportional to the cube of the orbit's semi-major axis.
Proof: T = 2πr/v = 2πr(r/GM)0.5 = 2π(r3/GM)0.5, T2=4π2r3/GM, which shows that T2 is proportional to r3. For elliptical orbiting, r = semi-major axis (a).
Since another variable is M which is the mass of star, the proportion constant 4π2/GM is applicable to all planets which orbits the same mass (that is being orbited) only.
Apparent weightlessness
We feel weight because there’s normal reaction act by ground on ourselves. i.e., R = W + ma.
The acceleration refers to the relative acceleration between you and the ground. When the relative acceleration between you and the ground is zero, you feel no weight since there's no normal reaction acting on you. (But actually you are still massive. Note that the mass won’t change.)
Some relating phenomenon:
-          When the balance is drop with the object on the balance, the readings is zero since there’s no force (mg) acting on the balance hence R = 0. The weight itself accelerates the object.
-          When astronauts in spaceship orbits a planet, the astronauts himself feels no weight because the spaceship has the same acceleration with him, hence no normal reaction acting on him.

First note for the Astronomy elective topic. I start with some calculation, then introducing the universe system and history, lastly celestial model and hardcore astron. knowledge.

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