The orbit of a planet is an ellipse with the sun at one focus. The sun’s gravitational force acts along the radial line from the planet to the sun (the dashed lines in Figure 15), and by Newton’s Second Law, the acceleration vector points in the same direction. Assuming that the orbit has positive eccentricity (the orbit is not a circle), explain why the planet must slow down in the upper half of the orbit (as it moves away from the sun) and speed up in the lower half. Kepler’s Second Law, discussed in the next section, is a precise version of this qualitative conclusion. Hint: Consider the decomposition of a into normal and tangential components.

Show that the car will not skid if the curvature κ of the road is such that (with R = 1/κ)

Note that braking (υ′ < 0) and speeding up (υ′ > 0) contribute equally to skidding.

Suppose that the maximum radius of curvature along a curved highway is R = 180 m. How fast can an automobile travel (at constant speed) along the highway without skidding if the coefficient of friction is μ = 0.5?

Beginning at rest, an automobile drives around a circular track of radius R = 300 m, accelerating at a rate of 0.3 m/s². After how many seconds will the car begin to skid if the coefficient of friction is μ = 0.6?

You want to reverse your direction in the shortest possible time by driving around a semicircular bend (Figure 16). If you travel at the maximum possible constant speed υ that will not cause skidding, is it faster to hug the inside curve (radius r) or the outside curb (radius R)? Hint: Use Eq. (5) to show that at maximum speed, the time required to drive around the semicircle is proportional to the square root of the radius.

What is the smallest radius R about which an automobile can turn without skidding at 100 km/h if μ = 0.75 (a typical value)?