Crackiitjee.in.Phy.ch15

For more Study Material and Latest Questions related to IIT-JEE visit www.crackiitjee.in

GRAVITATION Kepler’s Laws 1. Planets move in elliptical orbits with the sun at one focus. 2. As a planet moves in its orbit, a line drawn from the sun to the planet sweeps out equal areas in equal time intervals. 3. If T is the period and a is the length of a the semi-major axis, then T2/a3 is a constant

F g=



Gm 1 m 2 r2

G = 6.67 x 10-11 Nm2/kg2

.in

Newton’s Universal Law of Gravitation- any massive objects in the universe exert an attractive force on each other called the gravitational force.

itj

ee

Where m1 is the mass of the first object, m2 is the mass of the second object and r is the distance between their centers of masses. And Fg is the gravitational force between them. G is the universal gravitation constant.

ki

The force is negative because it is an attractive force.

cr

ac

Using this equation, you can determine the acceleration of gravity on the surface Gm 1 m 2 of a planet by setting m2g= and solving for g. r2 m2 will cancel out, m1 is the mass of the planet and r is the distance from the center of mass of the planet. If there are more than 2 objects, you can find the net force on one object by finding its gravitational attraction to each of the masses surrounding it and then performing a vector summation of those forces. Here it could be helpful to write the forces as unit vectors ( i and j). In the equation, the masses are directly proportional to the force. This means that if one mass is doubled, the force between the two objects is doubled. An important part of this equation comes from the fact that r 2 appears in the denominator. This is known as an inversed-square law.

For more Study Material and Question Bank visit www.crackiitjee.in

1

For more Study Material and Latest Questions related to IIT-JEE visit www.crackiitjee.in

Circular Orbits If we can assume a circular orbit for a satellite in orbit around a planet (or a planet around a star) we know that the gravitational force between them is causing the circular motion. In other words, centripetal force = gravitational force. Fcentripetal = Fgravitation m1v2/r=

Gm 1 m 2 r2

.in

where m1 is the mass of the satellite in orbit and m2 is the mass of the planet, and r is the distance between the objects’ centers.

ee

Using these equations, you can obviously solve for any of the variables included, but you can also solve for the satellite’s period. Period, T, is the time it takes for the satellite to complete one full revolution. 2 r v

cr

ac

ki

itj

T=

Gravitational Fields

Spherical Shell – When you are outside of the shell, it can be treated as if all the mass is at the center, a distance r away. When you are inside a spherical shell, you experience no net gravitational force. g(r) = 0 Fg = 0 for r< R g(r) =

Gm r2

Fg =

Gm 1 m 2 r2

for r  R

For more Study Material and Question Bank visit www.crackiitjee.in

2

For more Study Material and Latest Questions related to IIT-JEE visit www.crackiitjee.in

Gravitational Force vs. Distance for A Spherical Shell

r

ac

ki

itj

ee

.in

Fgravitation

R

cr

Solid Sphere – When you are outside of the sphere, it can be treated as if all the mass is at the center, a distance r away. When you are inside a solid sphere, the mass in the shell above you exerts no net gravitational force (like above). However, the mass in the sphere “below” you does exert a force. So when you are a distance r away from the center, you experience a gravitational force due to the mass and radius of the mass “beneath” you acting as its own planet. g(r) =

Gm R3

g(r) =

Gm r2

r

Fg =

Fg =

Gm 1 m 2 r R3 Gm 1 m 2 r2

for r< R

for r  R

For more Study Material and Question Bank visit www.crackiitjee.in

3

For more Study Material and Latest Questions related to IIT-JEE visit www.crackiitjee.in

The equations for r< R come from a more general equation Fg=

Gm 1 m in side r2

In this space, solve for

Fg =

Gm 1 m 2 r R3

using Newton’s Universal Law of gravitation. Hint: use density to put minside in terms of known variables such as the mass of the planet and the radius of the planet. Gravitational Force vs. Distance for A Solid Sphere

r

cr

ac

ki

itj

ee

Fgravitation

.in

R

For more Study Material and Question Bank visit www.crackiitjee.in

4

For more Study Material and Latest Questions related to IIT-JEE visit www.crackiitjee.in

Gravitational Potential Energy U = - -

U=-

Gm 1 m 2 dr r2 Gm 1 m 2 r

This means that the gravitational potential energy is negative at the surface of the planet and becomes closer and closer to zero until you are an infinite distance away, in which case the gravitational potential energy is zero.

e.

in

You will want to use this equation instead of U= mgh when you are dealing with a problem where the projectile is far from the surface of the Earth (when g is no longer 9.8 m/s2).

E= Uo + Ko

ck

Total mechanical energy

iit

je

Gravitational potential energy is also helpful when solving problems involving the conservation of mechanical energy, such as when a projectile is launched away from the surface of the Earth or when a projectile in space comes towards the Earth.

cr a

Uo + Ko = Uf + Kf

Instead of using mgh here, make sure you use the new form. -

Gm 1 m 2 Gm 1 m 2 + ½ m vo2 = + ½ m v f2 rf ro

(make sure that you use the distance from their center of masses for r).

For more Study Material and Question Bank visit www.crackiitjee.in

5

For more Study Material and Latest Questions related to IIT-JEE visit www.crackiitjee.in

Escape Speed- the minimum speed required to escape from the Earth’s gravity. If Ko > Uo then it will escape (unbounded) If Ko < Uo then it won’t escape (bounded) So if E = Ko + Uo  0, than the objects can escape! Gm 1 m 2 + ½ m vo2  0 ro

then

Gm 1 m 2 = ½ m vo2 ro 2Gmearth R

cr

ac

ki

itj

ee

so vescape =

.in

-

For more Study Material and Question Bank visit www.crackiitjee.in

6