Universal Law of Gravity · 2018-10-17 · Newton’s Law of Universal Gravitation G 6.67 10 11 N...

DepartmentofPhysicsandAppliedPhysicsPHYS.1410Lecture12A.Danylov

Lecture 12

Chapter 8

Universal Law of Gravity

Physics I

Course website:https://sites.uml.edu/andriy-danylov/teaching/physics-i/

He is right. I don’t feel my weight.But how to deal with anxiety

about falling?


Today we are going to discuss:Chapter 8:

Some leftovers Universal law of Gravity: Section 6.3 Circular Orbits/Satellites: Section 8.3

ConcepTest Ferris Wheel

A) N remains equal to mg

B) N is smaller than mg

C) N is larger than mg

D) none of the above

You’re on a Ferris wheel moving in avertical circle. When the Ferris wheel is atrest, the normal force N exerted by yourseat is equal to your weight mg. How doesN change at the top of the Ferris wheelwhen you are in motion?

You are in circular motion, so there has to be a

centripetal force pointing inward. At the top, the

only two forces are mg (down) and N (up), so N

must be smaller than mg.

Follow-up: Where is N larger than mg?

RvmNmg

2

NRvmmg

2

N

RvmmgN

2

RvmmgN

2

Bottom


Example Car on a circular flat road

R

vFc points toward the center of the circle (i.e., downward in this case). The weight vectorpoints down and the normal force (exerted by the hill) points up. The magnitude of the net force, therefore, isFc = mg – N.

N

Askiergoesoverasmallroundhillwith

radiusR.Becausesheisincircularmotion,

therehastobeacentripetalforce. Atthetop

ofthehill,whatisFc oftheskierequalto?

ConcepTest Going in Circles

Follow-up: What happens when the skier goes into a small dip?

A)Fc =N +mg

B)Fc =mg – N

C)Fc =T +N –mg

D)Fc =N

E)Fc =mg

Fc = N ‐mg

mg


Law of Universal GravitationThe motion of Satellites

Section 6.3Section 8.3


universal gravitational constant

M

mF

Gravitational force is central, attractive, proportional to masses, and inversely proportional to the square of the distance

2rmMGF

Newton’s Law of Universal Gravitation

G 6.671011 N m2 kg2

A gravitational force of attraction exists between all objects that have mass

Gravitationalforcesbetweentwoobjectsareonlysignificantwhenatleastoneoftheobjectsisverymassive.


Gravity near the Earth’s Surface

WhatshouldwegetfortheaccelerationduetogravityattheEarth’ssurface?

MEARTH 5.981024 kg REARTH 6.38106 m G 6.671011 Nm2

kg2

mg

g GME

RE2 9.80 m/s2

2rmGMF E

2rmGM

F E

ME

mLet’s apply N.2nd law for gravitational force

maF 2rmGMma E

Denote the acceleration due to gravity as 2)(

rGMrga E

If r=RE (on the surface)

F

So, we can simplify our force: That is why we use mg


Variations in “g” on the surface

The acceleration due to gravity varies over the Earth’s surface due to altitude, localgeology, and the shape of the Earth, which is not quite spherical.

TheaccelerationduetogravityvariesovertheEarth’ssurface

duetoaltitude anditissmalleronMt.Everest.

ConcepTest Mt. Everest/AppleTwo apples are droppedsimultaneously from 2m abovethe ground at the top of MountEverest and at sea level.Which hits the ground first?

A) Apple at sea level B) Apple on Mt. Everest C) They hit simultaneously

End of Class


Newton’s Cannon on a Mountain

https://physics.weber.edu/schroeder/software/NewtonsCannon.html


SatelliteExample

AsatellitedoesnotcrashintoEarthbecauseofitshighspeed.Ifitstoppedmoving,itwould,ofcourse,falldirectlyintoEarth.ThesamehappenswiththeMoon


Weighlessness (Explanation)

The “weightlessness” ina freely falling elevatoris because everything isaccelerating equallyindependent of mass.So, there is no normalforce.


The satellite and all its contents are in free fall, except with a huge tangential velocity. So, there is no normal force. Everything is accelerating equally independent of mass.This is what leads to the experience of weightlessness.

“Weightlessness” (cont.)

They do have a gravitational force acting on them, though!

https://www.gozerog.com/

Zero gravity created in a plane

WeightlessnesscanbeexperiencedonEarthaswell,butonlybriefly:


Thank you


mg

Tz

Tr

ar

T

z

r

)1(0 mgTF zz

)2(2

RmvSinTTr

rrr maTF

r component of tension provides the centripetal acceleration

Example Conical pendulum


Examples. Banked curve

But sometimes, friction force is not enough to keep a car on a circular road.

Banking the curve can help to keep cars from skidding.


mg

Ncos

Nsin

Fz = Ncos mg = 0

Fr = Nsin = marar= v2/R

N

Nsin mv2/R

tan v2/gRIndependent of object mass !!!

Banked Curves (solution)

r component of normal force provides the centripetal accelerationhttp://phys23p.sl.psu.edu/phys_anim/mech/car_banked_new.avi

r

z

Ncos mg

z

r

Take a ratio

Universal Law of Gravity · 2018-10-17 · Newton’s Law of Universal Gravitation G 6.67 10 11 N...

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