PHY 121/123 LECTURE NOTES - Week 4

During the first week we discussed motion with constant velocity, relative motion, and vectors. Next came costant acceleration, projectile motion, and vectors. Then we added forces, initially constant and therefore leading to constant or zero acceleration. This week we venture on, by looking at a special and important example of non-constant acceleration, where force is not necessarily changing in magnitude but it is changing in direction. We want to consider an object moving in a circle, which means that even if its speed around the circle is constant, it is accelerating, because the direction of its velocity keeps changing. The motion of the runner around a baseball diamond illustrates this. The average velocity or acceleration for a complete circuit is zero, but clearly there are accelerations as the runner goes around, always tending to push him in towards the center of the diamond. If after the start there were no acceleration, then the runner would just go to first base and continue in that direction forever.

When the motion is exactly in a circle at constant speed, our elementary calculus is enough to find the acceleration:

$${\bf\vec{x}}= (x_1, x_2) = r({\rm cos}\omega t,{\rm sin}\omega t)$$

$${\bf\vec{v}}=d{\bf\vec{x}}/dt = (dx_1/dt,dx_2/dt) = r\omega (-{\rm sin} \omega t,{\rm cos}\omega t)$$

$${\bf\vec{a}} = d{\bf\vec{v}}/dt = (dv_1/dt,dv_2/dt) = -r\omega^2 ({\rm cos}\omega t,{\rm sin}\omega t).$$

It is clear immediately that the direction of the acceleration is always opposite to the direction from the center of the circle to the position of the object. Further, the motion is at constant speed around the circle, with the angle being simply $\theta(t)= \omega t$, so that $\omega$, written in Latin letters as `omega', is the constant rate at which angle is increasing, and therefore is called circular frequency or angular velocity. The magnitude of the displacement from the center is simply the radius of the circle, $r=\sqrt{x_1^2 +x_2^2}$, while the magnitude of the velocity is $\vert{\bf\vec{v}}\vert=v=r\omega$ and the magnitude of the acceleration is $\vert{\bf\vec{a}}\vert=a=r\omega^2$. These last two relations give an expression for the acceleration in terms of the velocity and the radius of the circle,

a=v²/r .

This is a very powerful formula, with a wide variety of applications.

Let's start with some car and road situations. You are driving around a curve at 36 km/hr [about 22 mi/hr - not all that fast], and radius of the curve is 10 m [about 33 ft]. How much centripetal (i.e., towards the center) acceleration is required to keep you from running off the road? Now 3.6 km/hr is equivalent to 1 m/s, so this speed is 10 m/s, and now we have everything in standard units. Plugging into the magic formula above, we find the centripetal acceleration is 10 m/s², just as big as the downward acceleration of gravity. Clearly it would be dangerous to rely on friction between the tires and the road to supply this acceleration, because the friction force is less than or equal to $\mu mg$, and so can only supply an acceleration $\mu g$. As $\mu$ is almost always significantly smaller than unity, this won't be enough. Further, if anything makes the conditions slippery, then you'll go off the road for sure.

There is another way to keep you on the road, which is to `bank' it, that is, tilt the road so the inside of the curve is lower than the outside. If the road can be treated as a flat ribbon tilted towards the inside, with an angle $\theta$to the horizontal, then it is possible for the car to stay at fixed radius as it goes round the curve. Assuming no friction, this means that the normal force of the road on the car must supply both the horizontal centripetal force, which in this case is mg, and also the vertical upward force to balance the weight, also mg. As the normal force vector has equal vertical and horizontal components, it is tilted at 45^o to the vertical, and therefore the road surface must be tilted at the same angle to the horizontal. This sounds good, but it still is bad engineering. Suppose someone were driving very slowly around the curve, meaning much less centripetal force is needed. Then the car simply will slide sideways to the inside of the curve! Only on a professional racetrack could one even consider using such steep banking.

What then is the good answer to this challenge? If we want to allow speeds as high or higher than 36 km/hr, then the only way to reduce the required centripetal acceleration to manageable amounts is by increasing the radius of the curve. Because the acceleration goes as the square of the speed, to double the speed limit and keep things safe requires a curve with four times the radius. To get a feeling for the significance of the radius, it means that the road turns about 60^o when you go a distance equal to the radius. Next time you are on a high speed road, look at how much it curves, and you will see that the curves are very gentle. We can make this more quantitative. If the acceleration required to hold the car around the curve is to be less than 0.1 g, it should be feasible to manage with friction, banking, or more realistically, a combination of the two. For a speed limit of 100 km/hr [60 mi/hr], the radius of curvature for centripetal acceleration of 0.1g is about 770 m, or half a mile. Then if there were friction given by $\mu = 0.1$ and a bank slant of 5^o, any speed from zero up to almost 140 km/hr [85 mi/hr] would be safe. To take account of slippery conditions due to weather and the possibility of really excessive speeds, one probably still would want a safety margin of at least a factor two, hence a radius of one mile.

Newton's Theory of Gravity

Newton invented a picture of gravity which unified phenomena at very different distance scales. On the one hand, it agreed with the statement that near the surface of the earth there is a downward acceleration on any free object ${\bf \vec{a}}={\bf \vec{g}}= -g \hat{r}$, where $\hat{r}$ is a vector of unit magnitude pointing out from the center of the earth to the location of the object. On the other hand, it agreed also with the best information on the orbit of the moon around the earth and of planets around the sun.

That information is contained in three laws developed by Kepler, based on meticulous astronomical observations (unaided by a telescope!) of Tycho Brahe:

1. The orbit of a planet is an ellipse, with the sun at one focus.

2. A straight line from the sun to the planet sweeps out equal areas in equal times, meaning that the angle of the planet increases more slowly when it is far away.

3. The period T of an orbit for one revolution is proportional to $r\sqrt{r}$, where r is the radius of the orbit.

Newton showed, using more mathematics than we want to use, that all three laws would follow from his proposed universal law of gravitational attraction: The gravitational acceleration of one body (1) towards another body (2) is given by the rule

$${\bf \vec{a}}(r)=-Gm_2\hat{r}/r^2 ,$$

where r is the distance between the centers of the two bodies, $\hat{r}$ is a vector of unit magnitude in the direction from 1 to 2, and G is Newton's universal constant, $G=(2/3) \times 10^{-10}$ N m²/kg². Newton did not have enough information to determine the constant very accurately, but he got a quantitative verification of his identification between the force that pulls us down onto the earth with the force that holds the moon on its orbit around the earth.

Before reproducing this, let us verify that the third of Kepler's laws follows from Newton's theory, at least for circular orbits. It's very simple:

$$v^2/r = Gm_2/r^2 \Rightarrow (2\pi r/T)^2/r = Gm_2/r^2$$

or

$$T^2 = (2\pi)^2 r^3/Gm_2 \Rightarrow T = 2 \pi r \sqrt{r/Gm_2} .$$

It's worthwhile to check that the units work in this expression!

Now we are ready. For a satellite going around the equator on an orbit just above the surface of the earth, we have

v²/R_E = g .

Using g=10 m/s² and $R_E = 6.4 \times 10^6$ m, we get v=8,000 m/s = 8 km/s $\approx 29,000$ km/hr $\approx 17,000$ mi/hr - a wee bit above the speed limit in most places! The period of the motion, i.e., the time it takes to go once around the earth, is $T=2\pi R_E/ v \approx 1.4 hr$. [It's a good exercise to check this arithmetic.]

Next, we need to do the same thing for the moon, treating its orbit as a circle, which is not bad because the ellipticity is small, like that of most other planet and moon orbits. The key is the radius of the moon's orbit, $r_{moon} = 3.8 \times 10^8$ m $\approx 60 R_E$. This gives us a chance to express things in terms of a modern, beautiful, and useful idea, the gravitational field, which gives the gravitational acceleration at any distance from a center such as the earth:

g(r)=g (R_E/r)² .

Of course, the direction of the acceleration is towards the center of the earth. Technically, this can be described by writing the field as a vector,

$${\bf \vec{g}(\vec{r})} = -g\hat{r}(R_E/r)^2 ,$$

specifying explicitly the direction as well as the magnitude of the acceleration of any body at any distance due to its attraction towards the earth. Evidently the gravitational acceleration towards the earth of any object at the distance of the moon is $g(r_{moon})=g/(60)^2 \approx 2.8$ mm/s², a lot smaller than g on earth! By Kepler's third law the period of the moon's orbit around the earth should be $T_{moon} = 60\sqrt{60} T = 651$ hr $\approx 27$ days. When Newton got this result, he knew he was on the right track. Not only did his theory account for all of Kepler's conclusions about motions of planets around the sun, but also it related quantitatively the orbit of the moon to the trajectory of a projectile just above the surface of the earth. Here is an interesting question to ponder: While the moon goes round the earth, the earth also is going round the sun. How do we perceive a month to pass, between new moon and new moon? Is it the same as the time we just calculated for the moon to go in a closed orbit around the earth, or is it longer or shorter?

We can gain more insight on the Kepler-Newton discovery by looking at the geometry of the closed orbits. A circle can be described in a quite physical way by imagining a rope tied at one end to a pole, in such a way that the rope can swing freely around the pole. At the other end is a person who grasps the rope and leans back to keep the rope taut. Moving continually to the right, the person clearly goes in a circle, with the length of the rope being the radius. Now make the situation just a little more complicated, by introducing two poles, with one end of a long rope attached to each. Let a person lean against the rope, stretching it so that the rope is always taut. This time the path of the person describes an ellipse, which is more eccentric the larger the ratio between the separation of the poles and the length of the rope (which must be at least as big as the separation [Question: Why is this last statement true?]). By bringing the two poles together we recover the circle, obviously a special case of the ellipse. The two poles are the two foci of the ellipse, so called because if reflective material coated the inside of the ellipse, then light sent out from one focus would all be reflected to the other one (and back to the first, etc.).

Let's use this description in terms of elliptical orbits to analyze some interesting problems in Hecht's book, Ch. 5, multiple-choice questions 15-17.

15. Suppose the point on an elliptical orbit nearest to the focus S where the heavy planet or star sits is called A, and the farthest point is called C. These two points must lie on a straight line which also passes through both foci. The perpendicular line passing through S intersects the ellipse at B and D on opposite sides, and we are asked to determine the relation of satellite speeds at these four points on the ellipse. The answer is

$$v_{\rm A} \gt v_{\rm B} = v_{\rm D} \gt v_{\rm C} .$$

To explain this, notice that as the satellite goes from A to B to C it is getting farther from the star, and therefore the attractive force from the star opposes the motion, slowing it down. Clearly this reverses on the back side of the orbit from C to D to A, and that justifies the claimed results. The exact equality of speeds at B and D makes sense because of the symmetry of the figure - for example, if the motion were watched on a film which had been shown backwards, then the roles of B and D would be interchanged, but other than that the motion would look just the same.

16. A satellite is going in a circle, and at one point A on the orbit something is thrown out straight ahead. What sort of orbit will this object follow? We know it has to be an ellipse, and since the velocity at A is perpendicular to the radius, it is not going in or out at A. Therefore A must be the near point or the far point. It cannot be the far point because the new orbit clearly will lie outside the circle (gravity is not big enough to keep it going in the circle with its new, higher speed), so the other end will be farther away from the focus than this one. Therefore, A must be the near point. This is enough to sketch the orbit and recognize it as an ellipse entirely outside the circle of the satellite.

17. In this problem, a rocket is to be brought down from a circular orbit by two successive thrusts. If initially at A it is given an accelerating thrust, then like the thrown object in the last problem it will go on a new orbit with A as the near point and E as the far point. Because E is far away, the motion is slowest there, and so a small braking thrust will be enough to stop it dead, and allow it to fall straight in.