Central Forces: Circular Motion and Gravitation

Circular motion: object moving in a circle of radius r, with constant speed v. r T = period = time for 1 complete revolution, 1 cycle v ( Don't confuse tension T with period T.)

distance 2p r speed v= v = = time T

An object moving in a circle is accelerating, because its velocity is changing, -- changing direction. Recall the definition of acceleration:     Dv v - v a � 2 1 , velocity v can change is two ways: Dt D t

v Magnitude can change or v v direction can change: 1 v v 2 v 2 1

For circular motion with constant speed, one can show that  v2 1) the magnitude of the acceleration is a= a r

2) the direction of the acceleration is always towards the center of motion. This is centripetal acceleration. "centripetal" = "toward center" r Notice that the direction of acceleration vector is always changing, a v therefore this is not a case of constant acceleration (can't use the a "constant acceleration formulas") v Is claim (1) sensible? 骣m2 2 琪 2 Check units: 轾v2 (ms) 桫s m Yep. = = = = [a] 犏 2 臌r m m s

Think: to get a big a, we must have a rapidly changing velocity. Here, we need to rapidly change the direction of vector v  need to get around circle quickly  need either large speed v or a small radius r.  a = v2 / r makes sense. (Proof is given in the Appendix.)

Is claim (2) sensible? Observe that vector v is toward center of circle.

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v v 1 Direction of a = 2 direction toward which v velocity is changing 1 v v 2

Example: acceleration on a merry-go-round. Radius r = 5 m , period T = 3 s

2p r 2 p( 5 ) r v= = = 10. 5 m/s v T 3 a

v2 (10. 5)2 a == = 22 m/s2 = 2. 3 g's ! r 5

A human can withstand ~ 5 g's for a few minutes or ~10 g's for a few seconds without losing consciousness.

Forces and circular motion   NII: Fnet = m a  To make something accelerate, we need a force in the same direction as the acceleration.  Centripetal acceleration is always caused by centripetal force (force toward center).

Example: Rock twirled on a string. (Assume no gravity)

m = 0.1 kg , T (period) = 1 s , radius r = 1 m r F T What is tension FT in the string?

F is the only force acting. (No such thing as F T T "centrifugal force"!)

v2 2p r F= m a = m, v = , T r T 2 (2p r / T ) r 1 F= m = 4 p2 m = 4 p 2( 0 . 1 ) = 3 . 9 N 1 pound T r T2 1 2

What about the outward "centrifugal force"?

A person on a merry-go-round (or twirled on a rope by a giant) "feels" an outward force. This is an illusion! There is no outward force on the person. Our intuition is failing us. Our intuition about forces was developed over a lifetime of experiences in inertial (non-accelerating) reference frames. If we are suddenly placed in an accelerating reference frame, our brains (wrongly) interpret our sense impressions as if we were still in a non-accelerating frame.

The result is that the direction of the perceived force is exactly opposite the direction of the true force. Example: A person in car accelerating forward. The chair pushes the driver forward. The force on the driver is in the forward direction. But the driver "feels" herself pressed back into the seat. It seems there is some force pushing the driver backward. WRONG!

"Centrifugal force" (not to be confused with "centripetal force") is also called a "pseudo-force" or "fictitious force" . Newton's Laws are only valid in a non-accelerating reference frame (an inertial frame). If we try to analyze motion in a non-inertial frame (for instance, in a rotating frame) then Newton's Laws don't hold. However, we can pretend that Newton's Laws hold in an accelerating frame if we pretend that "pseudo-forces" exist. That is, we can get the right answer if we makes two mistakes. In my opinion, this is a Devil's bargain. Computational convenience has come at the price of endless confusion of millions of physics students (and many professional engineers!). My advice: If you have choice, NEVER do calculations in non-inertial frames. Avoid using fictitious forces.

Consider the rock on the string again (still no gravity). If the string breaks, then there is no longer any force on the rock and it will move in a straight line with constant velocity. [according to NI: if Fnet = 0, then v = constant]. The reason the rock does not move in a straight line is because the string keeps pulling it inward, turning it away from its straight-line path. There is NO outward force on the rock.

Example: Rotation with friction. A car rounds a curve on a flat road (not banked). The radius of the circular curve is r = 100 m, and the speed of the car is v = 30 m/s (  68 mph). How large a static friction coefficient (S, not K !) is needed for the car to not skid off the road?

r N v F a fric a mg F fric

Top view View from rear of car

2 Fnet = Ffric = m a  S N = m v / r (car about to skid  Ffric = S N )

2 N = mg  S m g = m v / r (m's cancel) 

v2( 30m/s ) 2 mS = =2 = 0. 92 (no units) g r( 9 . 8m/s )( 100m )

So, need S  0.92 or else car will skid. (For rubber on dry asphalt S  1.0 , for rubber on wet asphalt S  0.7. So car will skid if road wet.)

Gravity!

Newton's Universal Law of Gravitation (first stated by Newton): any two masses m1 and m2 exert an attractive gravitational force on each other according to

m1 m 2 F= G r2

m m 1 F F 2

G = universal constant of gravitation = 6.67  10–11 N m2 / kg2 (G is very small, so it is very difficult to measure!)

Don't confuse G with g.

Newton showed that the force of gravity must act according to this rule in order to produce the observed motions of the planets around the sun, of the moon around the earth, and of projectiles near the earth. He then had the great insight to realize that this same force acts between all masses. [That gravity acts between all masses, even small ones, was experimentally verified in 1798 by Cavendish.]

Newton couldn't say why gravity acted this way, only how. Einstein (1915) General Theory of Relativity, explained why gravity acted like this.

Example: Force of attraction between two humans. 2 people with masses m1  m2  70 kg a distance r = 1 m apart. -11 2 m1 m 2 (6 . 67 10 )( 70 ) F= G = = 3. 3 10-7 N r2 1 2

This is a very tiny force! It is the weight of a 3.4 10–5 gram mass. A hair weighs 210–3 grams – the force of gravity between two people talking is about 1/60 the weight of a single hair.

Computation of g

Important fact about the gravitational force from spherical masses: a spherical body exerts a gravitational force on surrounding bodies that is the same as if all the sphere's mass were concentrated at its center. This is difficult to prove (Newton worried about this for 20 years.)

test mass m

F grav

sphere, mass M test mass m

F (same as with sphere) point mass M grav

We can now compute the acceleration of gravity g ! (Before, g was experimentally determined, and it was a mystery why g was the same for all masses.)

Fgrav = m a = m g

Earth mass m, M m GE = m g dropped near 2 R E R surface E (since r = RE is distance from m to center of Earth)

mass M E

G ME m's cancel !  g = 2 R E

Newton's Theory explains why all objects near the Earth's surface fall with the same acceleration. Fnet = ma says that a bigger force is required to accelerate a bigger mass m. Fgrav = GMm/r2 says that a bigger mass m feels a bigger force. So near the earth, bigger masses experience a bigger force in a way that produces the same acceleration for all masses. Newton's theory also makes a quantitative prediction for the value of g, which is correct.

Example: g on Planet X. Planet X has the same mass as earth (MX = ME) but has ½ the radius (RX = 0.5 RE). What is gx , the acceleration of gravity on planet X?

Planet X is denser than earth, so expect gx larger than g.

G MX G M E1 G M E gx =2 =2 = 2 2 = 4 g RXR E (1/ 2) R E . Don't need values of G, ME, and RE! ( 2 ) 142 43 123 4 g of earth

Or set up a ratio:

骣G/ MX 琪 2 2 g桫R X M骣 R 2 x= = X E =1� 2 = 4, g 4 g G/ M 琪 X E gE骣 E M E桫 R X 琪 2 桫R E

______* ______

At height h above the surface of the earth, g is less, since we are further from the surface, further from the earth's center.

r = RE + h  h G ME G M E g =2 = 2 earth r( RE + h )

The space shuttle orbits earth at an altitude of about 200 mi  1.6 km/mi  320 km. Earth's radius RE = 6380 km. So the space shuttle is only about 5% further from the earth's center than we are. If r is 5% larger, then r2 is about 10% larger, and

ME m Fgrav( on mass m in shuttle)= G2 @ about 10% less than on earth's surface (RE + h )

Astronauts on the shuttle experience almost the same Fgrav as when on earth. So why do we say the astronauts are weightless??

"Weightless" does not mean "no weight". "Weightless" means "freefall" means the only force acting is gravity.

If you fall down an airless elevator shaft, you will feel exactly like the astronauts. You will be weightless, you will be in free-fall.

astronaut v An astronaut falls toward F N the earth, as she moves grav forward, just as a bullet fired horizontally from a F Earth grav gun falls toward earth.

Orbits Consider a planet like Earth, but with no air. Fire projectiles horizontally from a mountain top, with faster and faster initial speeds.

would go straight, if no gravity The orbit of a satellite around the earth, or of a planet around the sun Planet obeys Kepler's 3 Laws.

Kepler, German (1571-1630). Before Newton. Using observational data from Danish astronomer Tycho Brahe ("Bra-hay"), Kepler discovered that the orbits of the planets obey 3 rules.

orbits!

KI : A planet's orbit is an ellipse with the Sun at one focus. Planet

KII : A line drawn from planet P to sun S sweeps out equal Sun areas in equal times. same time intervals same areas S slower faster

KIII: For planets around the sun, the period T and the mean distance r from the sun are related T2 T2 T 2 A= B by 3 = constant . That is for any two planets A and B, 3 3 . This means that r rA r B planets further from the sun (larger r) have longer orbital periods (longer T). ______* ______

Kepler's Laws were empirical rules, based on observations of the motions of the planets in the sky. Kepler had no theory to explain these rules.

Newton (1642-1727) started with Kepler's Laws and NII (Fnet = ma) and deduced that

MS m P Fgrav = G 2 . Newton applied similar reasoning to the motion of the Earth-Moon (Sun- planet ) rSP

ME m system (and to an Earth-apple system) and deduced that Fgrav = G 2 . (Earth-mass m ) rEm Newton then made a mental leap, and realized that this law applied to any 2 masses, not just to the Sun-planet, the Earth-moon, and Earth-projectile systems.

2 Starting with Fnet = ma and Fgrav = G Mm / r , Newton was able to derive Kepler's Laws (and much more!). Newton could explain the motion of everything!

Derivation of KIII (for special case of circular orbits). Consider a small mass m in circular orbit about a large mass M, with orbital radius r and period T. We aim to show that T2 / r3 = const. Start with NII: Fnet = m a

period T The only force acting is gravity, and for circular motion M a = v2 / r 

2 r M m2 2p r / v M 2 骣 G2/ = m/ � G = v 琪 v r r/ r桫 T m [recall the v = dist / time = 2r / T ]

M 4p2 r 2 T 2 4 p 2 �G � = constant, independent of m r T2 r 3 G M (Deriving this result for elliptical orbits is much harder, but Newton did it.)

An extra result of this calculation is a formula for the speed v of a satellite in circular orbit: GM v = . For low-earth orbit (few hundred miles up), this orbital speed is about 7.8 km/s r  5 miles/second.

Measurement of Big G

The value of G ("big G") was not known until 1798. In that year, Henry Cavendish (English) measured the very tiny Fgrav between 2 lead spheres, using a device called a torsion balance.

2 m1 m 2 F grav r Fgrav = G2 G = ( If Fgrav, r, and m's known, can compute G.) r m1 m 2

G ME Before Cavendish's experiment, g and RE were known, so using g = 2 , one could R E compute the product GME, but G and ME could not be determined separately.

With Cavendish's measurement of G, one could then compute ME. Hence, Cavendish "weighed the earth".

Appendix: Proof of a = v2 / r for circular motion

The proof involves geometry (similar triangles). It is mathematically simple, but subtle.

Consider the motion of a particle on a circle of radius r with v r 2 constant speed v. And consider the position of the particle at two 2 times separately by a short time interval t. (In the end we will r take the limit as t  0.) We can draw a vector diagrams 1 v       1 representing r1+ D r = r 2 and v1+ D v = v 2 : v r 2 r = v t v r 2 v 1 1

Notice that these are similar triangles (same angles, same length ratios) Also, note that     r1= r 2 = r and v1= v 2 = v .

Dv D r vD t Because the triangles are similar, we can write = = . A little algebra gives v r r Dv v v v2 = , which is the same as a = . Dt r r

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