the mathematics of star trek lecture 3: equations of motion and escape velocity

The Mathematics of Star Trek

Lecture 3: Equations of Motion and Escape Velocity

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Topics

Antiderivatives Integration Differential Equations Equations of Motion Escape Velocity

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Antiderivatives

We’ve already seen the idea of the derivative of a function.

A related idea is the following: Given a function f(x), find a function F(x)

such that F’(x) = f(x). If such a function F(x) exists, we call F

an antiderivative of f.

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Antiderivatives (cont.)

For example, given f(x) = 3x2, an antiderivative of f is F(x) = x3, since F’(x) = 3x2.

Question: Can you think of any other antiderivatives of f(x) = 3x2?

Possible answers: G(x) = x3 + 1, H(x) = x3 + 4, K(x) = x3 - 5, etc.

Notice that all these antiderivatives of f(x) = 3x2 differ by a constant!

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Antiderivatives (cont.) Graphically, this should

make sense, since for a fixed x-value, all of the antiderivatives given above have the same slope!

This is true in general! If F(x) and G(x) are

antiderivatives of f(x) on an interval, then F(x) = G(x) + C, for some constant C.

-2 -1 0 1 2

-10

-5

0

5

10

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Integration

The process of finding an antiderivative of a given function f(x) is called antidifferentiation or integration.

If F’(x) = f(x), then we denote this by writing ∫ f(x) dx = F(x) + C.

We call the symbol ∫ an integral sign, f(x) the integrand, and C the constant of integration.

Thus, we can write ∫ 3x2 dx = x3 + C. This notation is due to Leibnitz!

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Integration (cont.)

Using the fact that integration is differentiation “backwards”, we can apply derivative shortcuts to get integration shortcuts!

∫ k f(x) dx = k ∫ f(x) dx for any constant k. ∫ [f(x) + g(x)] dx = ∫ f(x) dx + ∫ g(x) dx. ∫ xn dx = xn+1/(n+1) + C for any rational

number n -1. ∫ ek x dx = (1/k)ek x + C.

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Integration (cont.)

For practice, evaluate each integral! ∫ x4 dx ∫ 2u7 du ∫ (t2 + 3 t + 1) dt ∫ 3 e3 x dx

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Differential Equations

One branch of mathematics is differential equations.

Many applications that involve rates of change can be modeled with differential equations.

A differential equation is an equation involving one or more derivatives of an unknown function.

When solving a differential equation, the goal is to find the unknown function(s) that satisfy the given equation.

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Differential Equations (cont.)

Here are some differential equations:

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We can use integration to help solve certain differential equations!

A differential equation is said to be separable if it can be put into the form:

In this case, rewrite the equation in the form s 1/g(y) dy = s f(x) dx and integrate each side with respect to the appropriate variable.

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Here is an example! An object moving along a straight line with

under the influence of a constant acceleration a is described by the differential equation:

dv/dt = a, where v is the object’s velocity at time t. We can use separation of variables to solve for

velocity v!

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∫ dv = ∫ a dt v = a t + C (general solution) If the object has initial velocity v0 at time t

= 0, then we can find C. v0 = a (0) + C

v0 = C

Thus, v = a t + v0 (particular solution).

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Recall that velocity is the derivative of the object’s position function s(t).

It follows that the object’s position function satisfies the differential equation:

ds/dt = a t + v0. If the object has initial position s0 at time t = 0,

then separation of variables can be used to show that:

s = ½ a t2 + v0 t + s0. (HW: Show this!)

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Equations of Motion

In this last example, we have derived the Equations of Motion for an object moving along a straight line, under the influence of a constant acceleration a, with initial position s0 and initial velocity v0:

s = ½ a t2 + v0 t + s0

v = a t + v0.

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Equations of Motion (cont.) One application of these equations of motion is

projectile motion. For example, suppose Commander Sisko’s baseball is

thrown straight up into the air with an initial velocity of 30 m/sec.

Assuming that the acceleration due to gravity is -9.8 m/sec2, find each of the following:

(a) The time at which the ball reaches its maximum height.

(b) The maximum height that the ball reaches. (c) The time at which the ball hits the ground. (d) The velocity with which the ball hits the ground.

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Equations of Motion (cont.) Solution: (a) With initial height s0 = 0 m and initial velocity v0 =

30 m/sec, the ball’s equations of motion are: s = -4.9 t2 + 30 t + 0 (m) v = -9.8 t + 30 (m/sec) At the ball’s maximum height, the velocity is zero, so

using the second equation: 0 = -9.8 t + 30 t = 30/9.8 = 3.06 seconds is time at which maximum

height is reached.

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Equations of Motion (cont.)

(b) To find the maximum height the ball reaches, use the first equation with t = 30/9.8 sec:

s = -4.9 (30/9.8)2 + 30 (30/9.8) = 45.9 m.

Maximum height of the ball reaches is approximately 45.9 meters.

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(c) When the ball hits the ground, its height will be zero, so using the first equation,

0 = -4.9 t2 +30 t = t(-4.9 t + 30), Thus t = 0 sec or t = 30/4.9 = 6.1 sec. The ball hits the ground approximately

6.1 seconds after it is thrown into the air.

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(d) Using the equation for velocity, when the ball hits the ground it’s velocity will be:

v = -9.8 (30/4.9) +30 = -30 m/sec.

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Escape Velocity Another question that we

can use differential equations to answer is the following:

What initial velocity is required for an object to escape the Earth’s gravitational field?

To answer this question, we need Newton’s Law of Universal Gravitation and Newton’s Second Law of Motion!

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Escape Velocity (cont.)

Newton’s Law of Universal Gravitation: The gravitational force between two masses M and m is proportional to the product of the masses and inversely proportional to the square of the distance between them, i.e. F = GMm/r2, where G is a constant.

Newton’s Second Law: The net external force on an object is equal to its mass times acceleration, i.e. F = ma.

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Using these laws, we find that the acceleration of an object a distance r from the Earth’s center is given by the equation:

a = dv/dt = -k/r2, where k is a constant

of proportionality.

R

r

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When r = R, then a = -g, the acceleration at the surface of the Earth, so:

-g = -k/R2, which yields k = gR2. Thus, a = dv/dt = -gR2/r2. Now, since v is a function of position r and r is

a function of time t, we can write: dv/dt = dv/dr dr/rt = v dv/dt. (This is an

application of the Chain Rule from Calculus.)

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Substituting v dv/dr for dv/dt in the equation dv/dt = -gR2/r2, we are led to the following model for an object’s velocity as a function of distance from the Earth’s center:

v dv/dr = -gR2/r2. This differential equation can be solved

via separation of variables!

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∫ v dv = ∫ -gR2 r-2 dr ½ v2 = g R2 r-1 + C v2 = (2g R2)/r + 2 C (general solution) If the object leaving the Earth’s surface

has an initial velocity of v0, then we can find constant C!

v02 = (2g R2)/R + 2 C

C = ½ (v02 - 2g R)

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Escape Velocity (cont.) Thus, our solution to this differential equation is: v2 = (2g R2)/r + v0

2 - 2g R. In order for the velocity v to stay positive, we need

v02 - g R ≥ 0, which means that

We call the right-hand side of this last expression Earth’s escape velocity, i.e. the minimum initial velocity needed for an object to escape the Earth’s force of gravity.

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References

Calculus: Early Transcendentals (5th ed) by James Stewart

Elementary Differential Equations (8th ed) by Rainville, Rainville, and Bedient

Hyper Physics: http://hyperphysics.phy-astr.gsu.edu/hbase/hph.html

The Cartoon Guide to Physics by Larry Gonick and Art Huffman

the mathematics of star trek lecture 3: equations of motion and escape velocity

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