AND NEWTON’S LAW OF GRAVITATION

I. Speed and Velocity

Speed is distance divided by time…is it any different for an object moving around a circle?

The distance around a circle is C = 2πr, where r is the of the circle

So average speed must be the divided by the time to get around the circle once

C 2πr one trip around the circle is v = = known as the period. We’ll use T T big T to represent this time

SINCE THE SPEED INCREASES WITH RADIUS, CAN YOU VISUALIZE THAT IF YOU WERE SITTING ON A SPINNING DISK, YOU WOULD SPEED UP IF YOU MOVED CLOSER TO THE OUTER EDGE OF THE DISK? 1 These four dots each make one revolution around the disk in the same time, but the one on the edge goes the longest distance. It must be moving with a greater speed.

Remember velocity is a vector. The direction of velocity in circular motion is on a tangent to the circle. The direction of the vector is ALWAYS changing in circular motion.

2 II. ACCELERATION

If the velocity vector is always changing EQUATIONS: in circular motion,

THEN AN OBJECT IN CIRCULAR acceleration in circular MOTION IS ACCELERATING. motion can be written as, THE ACCELERATION VECTOR POINTS  v2 INWARD TO THE CENTER OF THE a = MOTION. r Pick two v points on the path.  4π 2r Subtract head to tail…the a = vf resultant is the change in v or 2 the acceleration vector T vi a a Assignment: check the units -vi and do the algebra to make

vf sure you believe these equations Note that the acceleration vector points in

3 III. Centripetal and Centrifugal If the acceleration vector points inward, there must be an inward force as well.

vf v a i

The INWARD FORCE on an object in circular motion is called CENTRIPETAL FORCE

Without centripetal force, an object could not maintain a circular motion path. The velocity vector is telling you the object wants to go straight.

Centripetal force acts in a direction perpendicular to the velocity vector (the force, as it must be, is in the same direction as the acceleration).

Centripetal force causes the acceleration NOT by increasing (or changing) speed but by changing direction 4 The outward force is fictitious. It is called CENTRIFUGAL force.

you in the you before turn curve you FAST merry-go-round car direction before curve

You feel like you are being Banking into a left curve, you feel pushed outward when you like you are being thrown by an are on a circular motion path. outward force to the right. This is only because your That’s not a real force. That’s you, moving body WANTS to wanting to continue in your travel tangentially to the straight line path (your inertia resists circular path. the change) meeting the side of the car that is moving in a circle. The only thing keeping you in a circle is YOU holding on. You stay in a circular path thanks to your seat belt and the closed door of the car. 5 III. Some Equations for Circular Motion

We know from Newton’s 2nd Law that F = ma

Here’s how the equation applies to circular motion

v2 F = m r Using these equations, you can solve problems involving force on objects moving in a circular path. 4π 2r F = m T 2

6 Example problem

A 945‐kg car makes a 180‐ turn with a speed of 10.0 m/s. The radius of the circle through which the car is turning is 25.0 m. Determine the force of fricon acng upon the car.

This problem is way easier than it sounds. You just need to realize what the source and direction of the friction force is.

A car “in a 180 degree turn” just means it is in a circular motion path.

   F F norm = F grav  norm F   25 m frict  F frict = F net 2 Fgrav v F = m frict r m (10 )2 s The frictional force points inward. It is the Ffrcit = 945kg centripetal force required to keep the car 25m on the road in a circle. Ffrict = 3780N 7 IV. NEWTON”S LAW OF UNIVERSAL GRAVITATION

• Newton applied his ideas about why objects fall to earth to planetary orbits

• Basically he hypothesized that the Moon falls around the Earth for the same reason an object (like an apple?) falls toward the Earth…an attractive force he called GRAVITY

path of projectile without gravity

Assuming no air resistance…

Newton reasoned that if one could launch a projectile with sufficient speed that its horizontal path followed the Earth’s curvature, the force of gravity would cause it to “fall around” the Earth. The projectile would always fall toward the Earth without ever striking it.

Launch speeds too low, as we know from experience, crash to the Earth due to gravity

According to Newton, the projectile falling around the Earth was an analogy for the Moon orbiting the Earth

8 • Newton knew that objects near the Earth accelerated by 9.8 m/s2 • He also knew the Moon accelerated toward the Earth by 0.00272 m/s2

• WHY THE DIFFERENCE IF GRAVITY CAUSES AN OBJECT TO FALL AND THE MOON TO ORBIT THE EARTH?

Now look at the ratio of the distances of Look at the ratio of the accelerations: from the surface of the Earth to its center and the Moon to the center of the Earth: m 9.8 2 1 s = 6378km 1 m = 0.00272 3600 382680km 60 s2

COMPARE THE TWO RATIOS ABOVE m 9.8 2 6378km 1 THIS SAYS THAT GRAVITY s = ( )2 = m 382680km 3600 DECREASES AS THE INVERSE 0.00272 SQUARE OF DISTANCE s2

9 NEWTON SHOWED:  1 F ∝ 2 r

Force of attraction between two objects is inversely proportional to the square of the distance Earth’s gravity force represented with separating them vectors. The arrows are smaller further away from Earth but still directed to the center of Earth

The Moon, being very far from the Earth, is less influenced by its gravity and so accelerates more slowly toward the Earth.

This not explain the source of the Moon’s velocity

10 Newton’s Law of Universal Gravitation is:

F F  m m 1 2 1 2 m m2 F grav = G 1 r2 r The force of attraction between two objects is

DIRECTLY PROPORTIONAL TO THE PRODUCT OF THEIR MASSES

INVERSELY PROPORTIONAL TO THE SQUARE OF THE DISTANCE BETWEEN THEM

The universal gravitational constant, G, was ingeniously determined by Cavendish in the 1700s. The value is use today is

m G = 6.67428 × 10−11 N( )2 kg

11 V. Computing gravitational acceleration

Using the Universal Gravitational equation, we can derive an equation that allows you to compute the acceleration of gravity for other objects (besides Earth)  m1mEarth F grav = G r2 m m m g = G 1 Earth 1 r2 m g = G Earth r2

mplanet r is the radius of the planet generally, g = G 2 r

12 VI. SATTELITE MOTION Remember the projectile falling around the Earth earlier in the notes? Remember the discussion of the Moon falling around the Earth?

This type of motion is the same as satellite motion and we can derive an equation for the velocity of a satellite orbiting the Earth using our equations for circular motion, centripetal force and universal gravitation.

A satellite in circular orbit, like a car rounding a curve MUST experience a centripetal force.  v2 …and the acceleration of the F net = msat the satellite is like the r acceleration due to  msat mEarth gravity equation on the F grav = G r2 last page…   F net = F grav mEarth v2 m m g = G m = G sat Earth r2 sat r r2 mEarth 2 mEarth v = G asat = G 2 r r velocity of satellite m orbiting Earth a v = G Earth r distance r from the center 13