Gravitation Test - 5

According to properties of gravitational force, gravitational force between the particles is independent of the presence or absence of other particles; so the principle of superposition is valid i.e. force on a particle due to number of particles is the resultant of forces due to individual particles i.e.

Kepler's 3rd Law: T2 = a3. Kepler's 3rd law is a mathematical formula. It means that if you know the period of a planet's orbit (T = how long it takes the planet to go around the Sun), then you can determine that planet's distance from the Sun (a is the length of the semimajor axis of the planet's orbit)

The orbit of a planet around the Sun (or of a satellite around a planet) is not a perfect circle. It is an ellipse—a “flattened” circle. The Sun (or the centre of the planet) occupies one focus of the ellipse. A focus is one of the two internal points that help determine the shape of an ellipse. The distance from one focus to any point on the ellipse and then back to the second focus is always the same.

Kepler law of areas describes the speed of a planet travelling in an elliptical orbit around the sun. A planet’s orbital speed changes, depending on how far it is from the Sun. The closer a planet is to the Sun, the stronger the Sun’s gravitational pull on it, and the faster the planet moves. The farther it is from the Sun, the weaker the Sun’s gravitational pull, and the slower it moves in its orbit. So, statement of Kepler's law of area is "The line joining the sun to the planet sweeps out equal areas in equal interval of time". i.e. areal velocity is constant.

Cavendish's apparatus for experimentally determining the value of G involved a light, rigid rod about 2-feet long. Two small lead spheres were attached to the ends of the rod and the rod was suspended by a thin wire. When the rod becomes twisted, the torsion of the wire begins to exert a torsional force that is proportional to the angle of rotation of the rod. The more twist of the wire, the more the system pushes backwards to restore itself towards the original position.

The gravitational force acting by a spherically symmetric shell upon a point mass inside it, is the vector sum of gravitational forces acted by each part of the shell, and this vector sum is equal to zero. That is, a mass m within a spherically symmetric shell of mass M will feel no net force

We know by Newton's law of gravitation, the force on the body of mass m, situated at height h to the surface of the earth of mass M_E is given by:

we also know F=mass X acceleration( here acceleration is acceleration due to gravity)

⇒ F=mg  -(2)

Equating (i) & (ii), we get

since gravitational force is attractive in nature, so if there are two particles of m₁ & m₂ .

So One Force will be on m1 which will be directed towards m 2. i.e 

And other force wiil be on m2 which will be directed towards m 1 i.e

Clearly, It can be seen that Because thsese forces are attracted to each other and direction of each force is opposite to other one

The value of G was experimentally determined by Lord Henry Cavendish using a torsion balance.Cavendish brought two large lead spheres near the smaller spheres attached to the rod. Since all masses attract, the large spheres exerted a gravitational force upon the smaller spheres and twisted the rod a measurable amount. Once the torsional force balanced the gravitational force, the rod and spheres came to rest and Cavendish was able to determine the gravitational force of attraction between the masses.Cavendish expressed his result in terms of the density of the Earth.

G=gRearth2Mearth=3g4πRearthρearth" role="presentation" style="display: inline; line-height: normal; word-spacing: normal; overflow-wrap: normal; white-space: nowrap; float: none; direction: ltr; max-width: none; max-height: none; min-width: 0px; min-height: 0px; border: 0px; padding: 0px; margin: 0px; position: relative;">G=gR2earthMearth=3g4πRearthρearth$G=g\frac{R_{earth}^2}{M_{earth}}=\frac{3g}{4\pi R_{earth}{\rho }_{earth}}$

After converting to SI units, Cavendish's value for the Earth's density, 5.448 g cm−3, gives

G = 6.74×10−11 m3 kg–1 s−2

Today, the currently accepted value is 6.67259 x 10-11 N m2/kg2.