6.1: Centripetal Forces
In the absence of a net force, objects move in a straight line. If they turn — that is, if their velocity changes, even only in direction — there must be an applied force. Forces which cause objects to turn around continuously in a circle are known as centripetal forces. When an object moves in a circle its velocity at any particular instant points in a direction tangent to the circle. The acceleration points towards the center of the circle, and so does the force acting on it. This is only natural, when you think about it — if you feel a force pushing you towards your left as you walk forward, you will walk in a circle, always turning left.
The reason the Earth orbits the sun is due to the Universal Law of Gravity and centripetal motion. The Universal Law of Gravity was a great achievement in the history of mankind. It unified the 'heavenly bodies' and the 'Earthly bodies'. Isaac Newton showed that an apple falling to the ground and the moon orbiting Earth are both explained by the same laws of physics.
Key Equations
Centripetal Force
\begin{align*} F_C = \frac{mv^2}{r} \begin{cases} m & \text{mass (in kilograms, kg)}\\ v & \text{speed (in meters per second, m/s}\text{)}\\ r & \text{radius of circle} \end{cases}\end{align*}
Gravitational Force
\begin{align*}F_G = \frac{Gm_1 m_2}{r^2} \begin{cases} m_1, m_2 & \text{Masses of objects}\\ G & \text{Gravitational Constant}, 6.67300 \times 10^{11} \text{m}^3 \text{kg}^{1} \text{s}^{2}\\ r & \text{Distance between objects} \end{cases}\end{align*}
Key Concepts
 An orbital period, \begin{align*}T\end{align*}, is the time it takes to make one complete rotation.
 If a particle travels a distance \begin{align*}2\pi r\end{align*} in an amount of time \begin{align*}T\end{align*}, then its speed is distance over time or \begin{align*}2 \pi r /T\end{align*}.
 An object moving in a circle has an instantaneous velocity vector tangential to the circle of its path. The force and acceleration vectors point to the center of the circle.
 Net force and acceleration always have the same direction.
 Centripetal acceleration is just the acceleration provided by centripetal forces.
 Any force can be a centripetal force. Centripetal force is the umbrella term given to any force that is acting perpendicular to the motion of an object and thus causing it to go in a circle.
 When undergoing centripetal motion you might 'feels' an outward pull, however this is not a true force, but merely the objects inertia. Remember, Newton's first law maintains that the natural state of an object is to go in a straight line at constant speed. Thus, when you make a right turn in your car and the basketball in the back seat flies to the left, that is because the car is moving right and the basketball is maintaining it's position and thus from your point of view moves to the left. Your point of view in this case is different from reality because you are in a rotating reference frame.
 The Geosynchronous orbit is the orbit where a satellite completes one orbit of the Earth every 24 hours, staying above the same spot (longitude) on Earth. This is a very important orbit for spy satellites and TV satellites among others. You force the speed of the satellite to be a value such that the satellite makes one rotation every 24 hours.
Universal Law of Gravity
In previous chapters we learned that gravity  near the surface of planets, at least  is a force that accelerates objects at a constant rate. At this point we can extend this description using the framework of Newton's Laws.
Newton's Laws apply to all forces; but when he developed them only one was known: gravity. Newton's major insight  and one of the greatest in the history of science  was that the same force that causes objects to fall when released is also responsible for keeping the planets in orbit. According to some sources, he realized this while taking a stroll through some gardens and witnessing a falling apple.
After considering the implications of this unification, Newton formulated the Law of Universal Gravitation: Any two objects in the universe, with masses \begin{align*} m_1 \end{align*} and \begin{align*} m_2 \end{align*} with their centers of mass at a distance \begin{align*} r \end{align*} apart will experience a force of mutual attraction along the line joining their centers of mass equal to:
\begin{align*}\vec{F_G}=\frac{Gm_1m_2}{r^2} && \text{Universal Law of Gravity, } \intertext{where G is the Gravitational constant:} G = 6.67300\times10^{11} \mathrm{m^3 kg^{1} s^{2}}\end{align*}
Here is an illustration of this law for two objects, for instance the earth and the sun:
Gravity on the Earth's Surface
In the chapter on force, we saw that the gravitational force or weight formula for objects near earth, \begin{align*}F_g = mg \end{align*}, is a special case of a more general result.
On the surface of a planet  such as earth  the \begin{align*} r \end{align*} in the Universal Law of Gravity is very close to the radius of the planet. A planet's center of mass is approximately at its center (for a spherical planet with azimuthal symmetry of density this is exactly true). In addition, we can safely ignore the increase in height off the surface of Earth. For instance, the earth's radius is about 6,000 km, while the heights we consider for this book are on the order of at most a few kilometers  so we can say to very good accuracy that for objects near the surface of the earth, the \begin{align*} r \end{align*} in the formula is constant and equal to the earth's radius. This allows us to say that gravity is constant on the surface of the earth. Here's an illustration:
For any object a height \begin{align*} h \end{align*} above the surface of the earth, the force of gravity may be expressed as:
\begin{align*}\vec{F_G} = \frac{Gm_{earth}m_{obj}}{(r_{earth}+h)^2} \intertext{Now we make the approximation that} r_{earth}+h \approx r_{earth} \intertext{then, we can rewrite the above as} \vec{F_G} = \underbrace{\frac{Gm_{earth}}{r_{earth}^2}}_{\vec{g_{earth}}}\times m_{obj} = m_{obj}\times{\vec{g}}&&\text{Gravity on Earth}\end{align*}
We can do this because the quantity in braces only has constants; we can combine them and call their product \begin{align*} g \end{align*}. Remember, this is an approximation that holds only when objects are near (think in our atmosphere) the surface of Earth.
We call the quantity \begin{align*} mg \end{align*} an object's weight. Weight is different from mass  which is identical everywhere  since it depends on the gravitational force an object experiences. In fact, weight is the magnitude of that force. To find the weight of an object on another planet, star, or moon, use the appropriate values in the Law of Gravity formula and solve for \begin{align*} F_G \end{align*}
Key Applications
 To find the maximum speed that a car can take a corner on a flat road without skidding out, set the force of friction equal to the centripetal force.
 To find the tension in the rope of a swinging pendulum, remember that it is the sum of the tension and gravity that produces a net upward centripetal force. A common mistake is just setting the centripetal force equal to the tension.
 To find the speed of a planet or satellite in an orbit, set the force of gravity equal to the centripetal force.
Centripetal Forces and Gravity Problem Set
 When you make a right turn at constant speed in your car what is the force that causes you (not the car) to change the direction of your velocity? Choose the best possible answer.
 Friction between your butt and the seat
 Inertia
 Air resistance
 Tension
 All of the above
 None of the above
 You buy new tires for your car in order to take turns a little faster (uh, not advised — always drive slowly). The new tires double your coefficient of friction with the road. With the old tires you could take a particular turn at a speed \begin{align*}v_o\end{align*}. What is the maximum speed you can now take the turn without skidding out?
 \begin{align*}4v_o\end{align*}
 \begin{align*} 2v_o\end{align*}
 \begin{align*}v_o\end{align*}
 \begin{align*}\sqrt{2v}_o\end{align*}
 Not enough information given
 A pendulum consisting of a rope with a ball attached at the end is swinging back and forth. As it swings downward to the right the ball is released at its lowest point. Decide which way the ball attached at the end of the string will go at the moment it is released.
 Straight upwards
 Straight downwards
 Directly right
 Directly left
 It will stop
 A ball is spiraling outward in the tube shown to the right. Which way will the ball go after it leaves the tube?
 Towards the top of the page
 Towards the bottom of the page
 Continue spiraling outward in the clockwise direction
 Continue in a circle with the radius equal to that of the spiral as it leaves the tube
 None of the above
 An object of mass \begin{align*}10 \;\mathrm{kg}\end{align*} is in a circular orbit of radius \begin{align*}10 \;\mathrm{m}\end{align*} at a velocity of \begin{align*}10 \;\mathrm{m/s}\end{align*}.
 Calculate the centripetal force (in \begin{align*}N\end{align*}) required to maintain this orbit.
 What is the acceleration of this object?
 Suppose you are spinning a child around in a circle by her arms. The radius of her orbit around you is \begin{align*}1\end{align*} meter. Her speed is \begin{align*}1 \;\mathrm{m/s}\end{align*}. Her mass is \begin{align*}25 \;\mathrm{kg}\end{align*}.
 What is the tension in your arms?
 In her arms?
 A racecar is traveling at a speed of \begin{align*}80.0 \;\mathrm{m/s}\end{align*} on a circular racetrack of radius \begin{align*}450 \;\mathrm{m}\end{align*}.
 What is its centripetal acceleration in \begin{align*}\;\mathrm{m/s}^2\end{align*}?
 What is the centripetal force on the racecar if its mass is \begin{align*}500 \;\mathrm{kg}\end{align*}?
 What provides the necessary centripetal force in this case?
 The radius of the Earth is \begin{align*}6380 \;\mathrm{km}\end{align*}. Calculate the velocity of a person standing at the equator due to the Earth’s 24 hour rotation. Calculate the centripetal acceleration of this person and express it as a fraction of the acceleration g due to gravity. Is there any danger of “flying off”?
 Neutron stars are the corpses of stars left over after supernova explosions. They are the size of a small city, but can spin several times per second. (Try to imagine this in your head.) Consider a neutron star of radius \begin{align*}10\;\mathrm{km}\end{align*} that spins with a period of \begin{align*}0.8\end{align*} seconds. Imagine a person is standing at the equator of this neutron star.
 Calculate the centripetal acceleration of this person and express it as a multiple of the acceleration \begin{align*}g\end{align*} due to gravity (on Earth).
 Now, find the minimum acceleration due to gravity that the neutron star must have in order to keep the person from flying off.
 Calculate the force of gravity between the Sun and the Earth. (The relevant data are included in Appendix \begin{align*}B\end{align*}.)
 Calculate the force of gravity between two human beings, assuming that each has a mass of \begin{align*}80 \;\mathrm{kg}\end{align*} and that they are standing \begin{align*}1 \;\mathrm{m}\end{align*} apart. Is this a large force?
 Prove g is approximately \begin{align*}10 \;\mathrm{m/s}^2\end{align*} on Earth by following these steps:
 Calculate the force of gravity between a falling object (for example an apple) and that of Earth. Use the symbol \begin{align*}m_o\end{align*} to represent the mass of the falling object.
 Now divide that force by the object’s mass to find the acceleration \begin{align*}g\end{align*} of the object.
 Our Milky Way galaxy is orbited by a few hundred “globular” clusters of stars, some of the most ancient objects in the universe. Globular cluster \begin{align*}\;\mathrm{M}13\end{align*} is orbiting at a distance of \begin{align*}26,000\end{align*} lightyears (one lightyear is \begin{align*}9.46\times 10^{15} \;\mathrm{m}\end{align*}) and has an orbital period of \begin{align*}220\end{align*} million years. The mass of the cluster is \begin{align*}10^6\end{align*} times the mass of the Sun.
 What is the amount of centripetal force required to keep this cluster in orbit?
 What is the source of this force?
 Based on this information, what is the mass of our galaxy? If you assume that the galaxy contains nothing, but Solarmass stars (each with an approximate mass of \begin{align*}2\times 10^{30} \;\mathrm{kg}\end{align*}), how many stars are in our galaxy?
 Calculate the centripetal acceleration of the Earth around the Sun.
 You are speeding around a turn of radius \begin{align*}30.0 \;\mathrm{m}\end{align*} at a constant speed of \begin{align*}15.0\;\mathrm{ m/s}.\end{align*}
 What is the minimum coefficient of friction µ between your car tires and the road necessary for you to retain control?
 Even if the road is terribly icy, you will still move in a circle because you are slamming into the walls. What centripetal forces must the walls exert on you if you do not lose speed? Assume \begin{align*}m = 650 \;\mathrm{kg}\end{align*}.
 Calculate the gravitational force that your pencil or pen pulls on you. Use the center of your chest as the center of mass (and thus the mark for the distance measurement) and estimate all masses and distances.
 If there were no other forces present, what would your acceleration be towards your pencil? Is this a large or small acceleration?
 Why, in fact, doesn’t your pencil accelerate towards you?
 A digital TV satellite is placed in geosynchronous orbit around Earth, so it is always in the same spot in the sky.
 Using the fact that the satellite will have the same period of revolution as Earth, calculate the radius of its orbit.
 What is the ratio of the radius of this orbit to the radius of the Earth?
 Draw a sketch, to scale, of the Earth and the orbit of this digital TV satellite.
 If the mass of the satellite were to double, would the radius of the satellite’s orbit be larger, smaller, or the same? Why?
 A top secret spy satellite is designed to orbit the Earth twice each day (i.e., twice as fast as the Earth’s rotation). What is the height of this orbit above the Earth’s surface?
 Two stars with masses \begin{align*}3.00\times10^{31} \;\mathrm{kg}\end{align*} and \begin{align*}7.00\times10^{30} \;\mathrm{kg}\end{align*} are orbiting each other under the influence of each other’s gravity. We want to send a satellite in between them to study their behavior. However, the satellite needs to be at a point where the gravitational forces from the two stars are equal. The distance between the two stars is \begin{align*}2.0\times10^{10} \;\mathrm{m}\end{align*}. Find the distance from the more massive star to where the satellite should be placed. (Hint: Distance from the satellite to one of the stars is the variable.)
 Calculate the mass of the Earth using only: (i) Newton’s Universal Law of Gravity; (ii) the MoonEarth distance (Appendix \begin{align*}B\end{align*}); and (iii) the fact that it takes the Moon 27 days to orbit the Earth.
 A student comes up to you and says, “I can visualize the force of tension, the force of friction, and the other forces, but I can’t visualize centripetal force.” As you know, a centripetal force must be provided by tension, friction, or some other “familiar” force. Write a two or three sentence explanation, in your own words, to help the confused student.
 A space station was established far from the gravitational field of Earth. Extended stays in zero gravity are not healthy for human beings. Thus, for the comfort of the astronauts, the station is rotated so that the astronauts feel there is an internal gravity. The rotation speed is such that the apparent acceleration of gravity is \begin{align*}9.8 \;\mathrm{m/s}^2\end{align*}. The direction of rotation is counterclockwise.
 If the radius of the station is \begin{align*}80 \;\mathrm{m}\end{align*}, what is its rotational speed, \begin{align*}v\end{align*}?
 Draw vectors representing the astronaut’s velocity and acceleration.
 Draw a free body diagram for the astronaut.
 Is the astronaut exerting a force on the space station? If so, calculate its magnitude. Her mass \begin{align*}m=65\;\mathrm{kg}\end{align*}.
 The astronaut drops a ball, which appears to accelerate to the ‘floor,’ (see picture) at \begin{align*}9.8 \;\mathrm{m/s}^2.\end{align*}
 Draw the velocity and acceleration vectors for the ball while it is in the air.
 What force(s) are acting on the ball while it is in the air?
 Draw the acceleration and velocity vectors after the ball hits the floor and comes to rest.
 What force(s) act on the ball after it hits the ground?
Answers to Selected Problems
 .
 .
 .
 .
 \begin{align*}100 \;\mathrm{N}\end{align*}
 \begin{align*}10 \;\mathrm{m/s}^2\end{align*}
 \begin{align*}25 \;\mathrm{N}\end{align*} towards her
 \begin{align*}25 \;\mathrm{N}\end{align*} towards you
 \begin{align*}14.2 \;\mathrm{m/s}^2\end{align*}
 \begin{align*}7.1 \times 10^3 \;\mathrm{N}\end{align*}
 friction between the tires and the road

\begin{align*}.0034\mathrm{g}\end{align*}
 \begin{align*}6.2 \times 10^5\;\mathrm{m/s}^2\end{align*}
 The same as a.
 \begin{align*}3.56 \times 10^{22}\mathrm{N}\end{align*}
 \begin{align*}4.2 \times 10^{7} \;\mathrm{N}\end{align*}; very small force

\begin{align*}g = 9.8 \;\mathrm{m/s}^2\end{align*}; you’ll get close to this number but not exactly due to some other small effects
 \begin{align*}4 \times 10^{26} \;\mathrm{N}\end{align*}
 gravity
 \begin{align*}2 \times 10^{41} \;\mathrm{kg}\end{align*}

\begin{align*}.006 \;\mathrm{m/s}^2\end{align*}
 \begin{align*}.765\end{align*}
 \begin{align*}4880 \;\mathrm{N}\end{align*}
 \begin{align*}\sim 10^{8} \;\mathrm{N}\end{align*} very small force
 Your pencil does not accelerate toward you because the frictional force on your pencil is much greater than this force.
 a. \begin{align*}4.23 \times 10^7\mathrm{m}\end{align*} b. \begin{align*}6.6 \ R_e\end{align*} d. The same, the radius is independent of mass
 \begin{align*}1.9 \times 10^7\mathrm{m}\end{align*}
 You get two answers for \begin{align*}r\end{align*}, one is outside of the two stars one is between them, that’s the one you want, \begin{align*}1.32 \times 10^{10}\mathrm{m}\end{align*} from the larger star.
 .
 .
 \begin{align*}v = 28\;\mathrm{m/s}\end{align*}
 \begin{align*}v\end{align*}down, \begin{align*}a\end{align*}right
 \begin{align*}f\end{align*}right
 Yes, \begin{align*}640\mathrm{N}\end{align*}
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