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Chapter 6: Q21 CQ (page 220)

Draw a free body diagram for a satellite in an elliptical orbit showing why its speed increases as it approaches its parent body and decreases as it moves away

Short Answer

Expert verified

The power of attraction between the satellite and the parent body is determined by the square of their distance.

Step by step solution

01

Definition of Satellite

A satellite is an item that is placed into orbit on purpose.

02

Determining why the speed of the satellite increases

The following is a free body diagram of a satellite in an elliptical orbit:

The centripetal force isFC, and the satellite's velocity isv.

The system's overall angular momentum is retained since there is no exterior force acting on it.

The centripetal force of attraction allows moving the satellite in an elliptical orbit. The power of attraction between the satellite and the parent body is determined by the square of their distance. The velocity of the satellite’s movement in an elliptical orbit increases as the distance between the parent body and satellite decreases and vice versa.

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Most popular questions from this chapter

On February 14, 2000 the NEAR spacecraft was successfully inserted into orbit around Eros, becoming the first artificial satellite of an asteroid. Construct a problem in which you determine the orbital speed for a satellite near Eros. You will need to find the mass of the asteroid and consider such things as a safe distance for the orbit. Although Eros is not spherical, calculate the acceleration due to gravity on its surface at a point an average distance from its center of mass. Your instructor may also wish to have you calculate the escape velocity from this point on Eros.

Semi-trailer trucks have an odometer on one hub of a trailer wheel. The hub is weighted so that it does not rotate, but it contains gears to count the number of wheel revolutions—it then calculates the distance travelled. If the wheel has a 1.15 m diameter and goes through 200,000 rotations, how many kilometres should the odometer read?

Modern roller coasters have vertical loops like the one shown in Figure. The radius of curvature is smaller at the top than on the sides so that the downward centripetal acceleration at the top will be greater than the acceleration due to gravity, keeping the passengers pressed firmly into their seats. What is the speed of the roller coaster at the top of the loop if the radius of curvature there is \({\bf{15}}.{\bf{0}}{\rm{ }}{\bf{m}}\) and the downward acceleration of the car is \({\bf{1}}.{\bf{50g}}\)?.

A number of amusement parks have rides that make vertical loops like the one shown in Figure. For safety, the cars are attached to the rails in such a way that they cannot fall off. If the car goes over the top at just the right speed, gravity alone will supply the centripetal force. What other force acts and what is its direction if: (a) The car goes over the top at faster than this speed? (b)The car goes over the top at slower than this speed?

Part of riding a bicycle involves leaning at the correct angle when making a turn, as seen in Figure. To be stable, the force exerted by the ground must be on a line going through the center of gravity. The force on the bicycle wheel can be resolved into two perpendicular components—friction parallel to the road (this must supply the centripetal force), and the vertical normal force (which must equal the system’s weight).

(a) Show that\(\theta \)(as defined in the figure) is related to the speed v and radius of curvature r of the turn in the same way as for an ideally banked roadway—that is,\(\theta = {\tan ^{ - 1}}\,{v^2}/rg\)

(b) Calculate \(\theta \) for a \(12.0{\rm{ m}}/{\rm{s}}\) turn of radius \(30.0{\rm{ m}}\) (as in a race).

Figure 6.36 A bicyclist negotiating a turn on level ground must lean at the correct angle—the ability to do this becomes instinctive. The force of the ground on the wheel needs to be on a line through the center of gravity. The net external force on the system is the centripetal force. The vertical component of the force on the wheel cancels the weight of the system while its horizontal component must supply the centripetal force. This process produces a relationship among the angle \(\theta \), the speed \(v\), and the radius of curvature \(r\) of the turn similar to that for the ideal banking of roadways.
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