Laws of Newton -
Kepler's Second Law
This lecture continues studying movement of objects in a central field. The familiarity with material presented in the lectures Central Force Field and Kepler's First Law is essential for understanding this educational material.
Kepler's Second Law states that a segment, connecting our Sun with any planet moving around a Sun, sweeps out equal areas during equal intervals of time.
As in the case of the Kepler's First Law, this Second Law has been based on numerous experiments and years of observation.
This Law can be formulated more mathematically.
Imagine a three-dimensional space with a single source of gravitation - a point-mass M at point O.
Some object comes into this field - a point-mass m. Its position at time t is at point P(t) and its velocity is v(t).
At time t=0 the position of our object is P(0) and velocity vector v(0).
As we know from previous lectures of Laws of Newton part of this course, the trajectory of our object will lie in the plane defined by vectors of initial position OP(0)=r(0) and initial velocity v(0) at time t=0.
Therefore, we can restrict our analysis to a two-dimensional case of trajectory lying completely within the plane defined by r(0) and v(0).
Let's choose a system of polar coordinates in this plane with an origin at point O, where the source of gravity is located, and a base axis defined by direction from point O to a position of our object at time t=0 - point P(0).
If our object during a time interval from t1 to t2 moved from point P(t1) to P(t2), its position vector r(t) swept up a sector bounded by r(t1), r(t2) and a trajectory from P(t1) to P(t2).
Let's introduce a function A(t) that represents an area of a sector bounded by r(0), r(t) and a trajectory from P(0) to P(t).
Then the area swept by position vector r(t) during the object's motion from time t1 to t2 equals
ΔA[t1,t2] = A(t2) − A(t1)
Using the above symbols, the Kepler's Second Law can be formulated as
If t2−t1 = t4−t3 then
A(t2)−A(t1) = A(t4)−A(t3)
The above condition is equivalent to a statement that
dA(t)/dt is constant.
Indeed, let t1 and t3 be any two moments of time and t2=t1+Δt and t4=t3+Δt.
Then t2−t1=Δt and t4−t3=Δt
Therefore,
A(t1+Δt)−A(t1) =
= A(t3+Δt)−A(t3)
Dividing both sides by Δt, we get
[A(t1+Δt)−A(t1)]/Δt =
= [A(t3+Δt)−A(t3)]/Δt
Taking this to the limit, when Δt→0, we get
dA(t)/dt|t=t1 = dA(t)/dt|t=t3
Since values t1 and t3 are chosen freely, it means that the derivative of a function A(t) is constant.
Hence, we conclude, A(t) is a linear function of time t.
In reverse, if we assume that the first derivative of function A(t) is constant and, therefore, A(t) is a linear function of time t, we can easily prove that
if t2−t1 = t4−t3 then
A(t2)−A(t1) = A(t4)−A(t3)
The constant first derivative of function A(t), that represents an area swiped by a position vector during the time from t=0 to some value t, is just a mathematical way of stating the Kepler's Second Law.
Proving this characteristic of function A(t) is a proof of the Kepler's Second Law.
Let's prove it then.
Assume that an object position vector during time interval Δt moved from r(t) to r(t+Δt).
These two vectors form a triangle whose area approximately equal to an area A(t+Δt)−A(t) of a sector swiped up by vector r(t) during the time Δt. The approximation will be better, as interval of time Δt tends to zero.
The third side of this triangle is a vector connecting the end points of position vectors.
An approximation of this vector's magnitude is a magnitude of velocity vector at time t multiplied by time interval Δt:
r(t+Δt) − r(t) ≅ v(t)·Δt
The area of a triangle formed by two vectors a and b equals to the half of a magnitude of a vector product of these vectors because(i) the area of a triangle with two sides a and b with an angle ∠φ between them is equal to
½·a·ha = ½·a·b·sin(φ)
where ha is an altitude onto side a
(ii) from the definition of a vector product
|a⨯b| = |a|·|b|·sin(φ)
Using this, we can say that an area of a triangle formed by r(t), r(t+Δt) and v(t)·Δt equals to ½|r(t)⨯v(t)|·Δt.
Recall that Angular Momentum of an object of mass m moving in some field, having a position vector r(t) and velocity v(t), is defined as
L(t) = m·r(t)⨯v(t).
But if the field is central, the Angular Momentum is a constant because a central force has no torque.
Therefore,
|r(t)⨯v(t)| = |L|/m is a constant.
An immediate consequence from this is that the area of a triangle formed by r(t), r(t+Δt) and v(t)·Δt is ½|L|/m·Δt.
When Δt→0, the area of our infinitesimal triangle tends to the area of a sector swiped up by position vector r(t) during infinitesimal time interval Δt ΔA(t)=A(t+Δt)−A(t).
Therefore,
limΔt→0ΔA(t)/Δt = ½|L|/m
which is a constant.
The limit above is a derivative of A(t) by time. Since it is a constant, A(t) is a linear function of time.
That proves the Kepler's Second Law.
No comments:
Post a Comment