5 Earth gravitation g and lift .ppt

Gravitation of earth

Introduction to rotational motion

Centrifugal Force due to rotationGravitational Force due to earthOrbital period (T) = Time required to complete one rotationAngular velocity = 2/T radian/ secEarth

Vc=Centrifugal forceGravitational force

Escape velocity

Escape velocityIf we throw a object up such a velocity that it goes out of earths gravity it velocity is called escape velocityIt means that by the time its velocity reduces to zero due to gravity it had reached a point where earths gravitational force is also is zero. Hence the object will never return to earth

Me = mass of earthR UThrow a ball of (mass m) up Initial velocity at point A = U, and Gravitational force = FgFinal velocity at point B, V=0 & gravitational force Fg =0 FgV=0Fg=0BAv = u +(- gt )

Me = mass of earthR UThus earth will not attract the object & it will not return back to earth The initial velocity U is called escape velocityFgV=0Fg=0BAv = u +(- gt )R

Ex :Calculate initial velocity such that ball we reach height 5 m Me = mass of earthR ABWhen we throw a ball up it KE gets converted to PEKE at point A = PE at point BPE at point B = m x 10 x 5 J KE at point A = x m x U2 J x m x U2 = m x 10 x 5 U2 = 2 x 10 x 5 U2 = 100 U = 10m/sec5mU m/sec

Calculations of escape velocity Me = mass of earthR ABTo throw object from point A on earth to a point B in space we have to do work against earth gravitational force FgThis can be calculated by formula work = F x distance moved

Escape velocity dr

rR Fg is continuously reducing as r increasesConsider a very small distance dr. at distance r from earth center We assume that over this distance Fg is constant as dr is very small Work done to lift ball from A to B dw = Fg x drABFg

Escape velocity Me = mass of earthdr

rR AAB Work done to lift ball from A to B dw = Fg x drIf we can add work done over every such small dr from A to B, we will get net work done against gravity. NowABFg

Calculations of work done Me = mass of earthdr

rUsing calculus we can add such small work( dw1+ dw2 +dw3+ -----)1 for r changing from A to B .(B is point at infinity where Fg =0 ).This give total work done. Ans is R ABPE at B = Work done from A to B(G ) ( Me m )R=BAJ

MeR ABPE at B = Work done from A to B = KE at AU

Value of gravitational force/ acceleration os different objects (Density is assumed to be uniform)g at equator & poleFg inside & outside a solid sphereFg inside & outside a shellg at height h above earthLift problemsDrag force

Earth radius at equator is more than poles g at equator = 9.780 m/sec2 g at pole = 9.832 m/sec2Weight of body as measured by spring balance(force due to earths gravitation) increases from equator to pole but mass remains same6.378 10 6 meters at equator6.351 10 6 meters at pole

Finding latitude from value of g

Value of gravitational force/ acceleration os different objects (Density is assumed to be uniform)g at equator & poleFg inside & outside a solid sphereFg inside & outside a shellg at height h above earthLift problemsDrag force

Gravitational force inside & outside a solid sphere

Case 1. If a particle of mass m is located outside a homogeneous solid sphere ofCase 2. If a particle of mass m is located inside a homogeneous solid sphere of

The sphere attracts mass out side if as though the mass of the sphere were concentrated at its center..r( R+ r )2

rFgM = mass of inner sphereThe net gravitational force at poet Q due to outer material of main sphere is zeror

Inner sphereThe sphere has a uniform density, , in kg/m3, Let the mass of the full sphere (radius R ) = M kgLet mass of the inner sphere (radius r = M kgInner sphereF

The sphere has a uniform density, , in kg/m3, Let the mass of the full sphere = M kgLet mass of the inner sphere = M kgInner sphereFInner sphereF

The sphere has a uniform density, , in kg/m3, Let the mass of the interior sphere = M kgLet mass of the inner sphere = M kgInner sphereFInner sphereFAs r reduces F will reduceAt center of sphere r = 0 hence F = 0 Gravitational force at center of solid sphere is = 0F

Graph of F for solid sphere

Fg = max, V = 0+veFg = 0 V = maxFg = - max V = 0Fg

Time for oscillation of a ball in tunnel = 84 .3 minTime for a satellite to make full round on surface of earth will be also 84.3min

A proposal was made to operate a transport system in a tunnel between two cities A & B, using thisOne-way trip well about 42 min. Precise calculation were done as Earths density is not uniform.Practical problems wereHow to make frictionless tunnel,Temperature inside earth is high Magma will come out etcAB

Value of gravitational force/ acceleration os different objects (Density is assumed to be uniform)g at equator & poleFg inside & outside a solid sphereFg inside & outside a shellg at height h above earthLift problemsDrag force

Gravitational force due to Spherical Shell of uniform density1 Particle located outside a spherical shell2. If the particle is located inside the shellCase 1Case 2

If a particle of mass m is located outside a spherical shell of mass M at point P, the shell attracts the particle as if the mass of the shell is concentrated at its centerrShell ofmass M

If the particle is located inside the shell (at point P in Fig ),the gravitational force acting on it = 0Shell of mass

Value of gravitational force/ acceleration os different objects (Density is assumed to be uniform)g at equator & poleFg inside & outside a solid sphereFg inside & outside a shellg at height h above earthLift problemsDrag force

The g decreases with altitude:h

Radius of earth = 6370kmOnce the altitude becomes comparable to the radius of the Earth, the decrease in the acceleration of gravity is much larger:

What is the acceleration due to gravity of an object at the altitude of the space shuttles orbit, about 400 km above the Earths surface?

If altitude of International Space Station is 385 km above the Earths surface, what is its period of rotation

Elevator & weightlessness problems25 june

Apparent WeightYour perception of your weight is based on the contact forces between your body and your surroundings. If your surroundings are accelerating, your apparent weight may be more or less than your actual weight.

Actual weight of a person is determined by his mass and the acceleration of gravity,The sensation of apparent weight comes from the fact that we are supported from the floor, chair, etc.When body is in "free fall (accelerating downward at the acceleration of gravity) then the body is not being supported.Then we fill weightlessUnderstanding perception of weight

The feeling of "weightlessness" occurs whenThere is no force of support on your body.This can be achieved in several ways .ExFree fall as there is no support.

In a elevator if cable breaks airplane coming down with acc =g

Satellite rotating around earth

Different sensations of apparent weight can occur in elevator when it moves withZero acceleration (zero or constant speed): No change in apparent weightAccelerates upward: Apparent wt. increases Accelerates downward: Apparent wt. reduces If the elevator cable breaks (free fall with downward acceleration = g.): Weightlessness is felt

Accn = 0Velocity = 0 or constantNo change in apparent weight

Floor support = Reaction of weight + support force due to upward acceleration Hence apparent weight is moreWhen accelerating upwards

When accelerating downwardsFloor support = Reaction of weight - support to accelerate you downwardHence apparent weight is less

F support = 0 Apparent weight = 0Elevator cable broken

Terminal velocity during free fall in air: When net force is = 0 , downward velocity remain constant, called terminal velocity

Controlling terminal velocity by parachute

Earth gravitational acceleration g does note depend on mass of objectHence an elephant & a feather will take same time to reach ground if there in no resistance

End

Reference slides

Geostationary Satellites for Meteorology (and Volcanology!)http://www.ssec.wisc.edu/data/geo/http://www.rap.ucar.edu/weather/satellite/http://www.ssec.wisc.edu/data/volcano.html

Orbital Energetics The equation K = -Ug is called The Virial Theorem. In effect, it says that for a planet in orbit around the Sun, if you turned its velocity by 90o, so that it pointed straight out of the Solar System, you would have only half the kinetic energy needed to escape the Suns gravity well.

Example:The Total Energy of a Satellite Show that the total energy of a satellite in a circular orbit around the Earth is half of its gravitational potential energy.Although derived for this particular case, this is a general result, and is called the Virial Theorem. The factor of is a consequence of the inverse square law.

g for a Solid SphereGravitational field of a solid uniform sphere

g for a Hollow SphereGravitational field of a uniform spherical shell

Tides Tidal forces can result in orbital locking, where the moon always has the same face towards the planet as does Earths Moon.

If a moon gets too close to a large planet, the tidal forces can be strong enough to tear the moon apart. This occurs inside the Roche limit; closer to the planet we have rings, not moons.

Tides This figure illustrates a general tidal force on the left, and the result of lunar tidal forces on the Earth on the right.

Example:The Orbiting Space Station You are trying to view the International Space Station (ISS), which travels in a roughly circular orbit about the Earth. If its altitude is 385 km above the Earths surface, how long do you have to wait between sightings?

According to stationary observerF = maTaking up as +ve R - mg = maR = m(g + a)If a = 0 ==> R = mg normal weightIf a is +ve ==> R = m(g + a) weight increaseIf a is -ve ==> R = m(g - a) weight decreaseMeasured weight in an accelerating Reference Frame

According to travellerF = ma R - mg = ma BUT in his ref. frame a = 0! so R = mg!!How come he still sees R changing as lift accelerates?Only if it is an inertial frame of reference! The accelerating lift is NOT!

Didnt we say the laws of physics do not depend on the frame of reference?

The radius of the Moons orbit is RM=3.84x108 m. If T = 2p[r/g] and g=9.81 m/s2, then the Moons orbital period should be TM = 2p[RM/g] = 2p[(3.84x108 m)/(9.81 m/s2)] = 3.93 x 104 s = 11 hr. However, the actual orbital period of the Moon is about 27.3 days = 2.36 x 106 s. How could this calculation be so badly off? (Weaker gravity?) Lets use the Moons orbital period and calculate gM, the acceleration due to Earths gravity at the orbit of the Moon. gM = RM(2p/T)2 = (3.84x108 m)[2p/(2.36x106 s)]2 = 2.72 x 10-3 m/s2 But an apple falls at gE = 9.81 m/s2. So lets try something. Well calculate the product gR2 for an apple at the Earths surface and for the Moon in orbit: gMRM2=(2.72x10-3 m/s2)(3.84x108 m)2 = 4.01x1014 m3/s2 gERE2 = (9.81 m/s2)(6.37x106 m)2 = 3.98x1014 m3/s2 These products are essentially equal, because gravity falls off ~ 1/R2. The same gravitational force law affects the apple and the Moon.The Apple& the Moon

The radius of the Moons orbit is RM=3.84x108 m. If T = 2p[r/g] and g=9.81 m/s2, then the Moons orbital period should be TM = 2p[RM/g] = 2p[(3.84x108 m)/(9.81 m/s2)] = 3.93 x 104 s = 11 hr. However, the actual orbital period of the Moon is about 27.3 days = 2.36 x 106 s. How could this calculation be so badly off? (Weaker gravity?) Lets use the Moons orbital period and calculate gM, the acceleration due to Earths gravity at the orbit of the Moon. gM = RM(2p/T)2 = (3.84x108 m)[2p/(2.36x106 s)]2 = 2.72 x 10-3 m/s2 But an apple falls at gE = 9.81 m/s2. So lets try something. Well calculate the product gR2 for an apple at the Earths surface and for the Moon in orbit: gMRM2=(2.72x10-3 m/s2)(3.84x108 m)2 = 4.01x1014 m3/s2 gERE2 = (9.81 m/s2)(6.37x106 m)2 = 3.98x1014 m3/s2 These products are essentially equal, because gravity falls off ~ 1/R2. The same gravitational force law affects the apple and the Moon.The Apple& the Moon

Example:The Total Energy of a Satellite Show that the total energy of a satellite in a circular orbit around the Earth is half of its gravitational potential energy.Although derived for this particular case, this is a general result, and is called the Virial Theorem. The factor of is a consequence of the inverse square law.

Gravitational Lensing Light will be bent by any gravitational field; this can be seen when we view a distant galaxy beyond a closer galaxy cluster. This is called gravitational lensing, and many examples have been found.

Gravitational Lensing

Tides Usually we can treat planets, moons, and stars as though they were point objects, but in fact they are not. When two large objects exert gravitational forces on each other, the force on the near side is larger than the force on the far side, because the near side is closer to the other object. This difference in gravitational force across an object due to its size is called a tidal force.

Tides Tidal forces can result in orbital locking, where the moon always has the same face towards the planet as does Earths Moon.

If a moon gets too close to a large planet, the tidal forces can be strong enough to tear the moon apart. This occurs inside the Roche limit; closer to the planet we have rings, not moons.

Tides This figure illustrates a general tidal force on the left, and the result of lunar tidal forces on the Earth on the right.

Sample problem 5.11