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Newton's Conclusions From Kepler's Laws

Kepler's laws and Newton's laws taken together imply that the force that holds the planets in their orbits by continuously changing the planet's velocity so that it follows an elliptical path is (1) directed toward the Sun from the planet, (2) is proportional to the product of masses for the Sun and planet, and (3) is inversely proportional to the square of the planet-Sun separation. This is precisely the form of the gravitational force, with the universal gravitational constant G as the constant of proportionality. Thus, Newton's laws of motion, with a gravitational force used in the 2nd Law, imply Kepler's Laws, and the planets obey the same laws of motion as objects on the surface of the Earth.

Definition

Definition of Binding Energy

Definition: A gravitational binding energy is the minimum energy that must be added to a system for the system to cease being in a gravitationally bound state. A gravitationally bound system has a lower (i.e., more negative) gravitational potential energy than the sum of its parts this is what keeps the system aggregated in accordance with the minimum total potential energy principle.

Definition

Define Escape Velocity

Escape Velocity:
The minimum velocity with which a body should be projected from the surface of the earth so that it escapes from the earth's gravitational field, is called the escape velocity

(V_{e})

of the body.

Formula

Expression for Escape velocity

Consider a satellite of mass m, stationary on the surface of earth, then binding of the satellite on the surface of earth is given by

B . E . = \frac{G M m}{R}

R - radius of earth
To escape from the surface of earth, the Kinetic energy of satellite equivalent to binding energy.
K.E. = B.E.

\frac{1}{2} m {v_{e}}^{2} = \frac{G M m}{R}

v_{e} = \sqrt{\frac{2 G M}{R}}

..........(1)
We know that

g = \frac{G M}{R^{2}}

hence,

v_{e} = \sqrt{2 g R}

..........(2)

Example

Find the variation of escape velocity

Escape velocity is a function of radius and gravitational field of the planet. Its value varies from planet to planet as radius and gravity changes.

Example: The ratio of the radius of a planet A to that of planet B is

r

. The ratio of accelerations due to gravity for the two planets is

x

. Find the ratio of the escape velocities from the two planets?

Solution:Escape Velocity is given by

V_{e} = \sqrt{2 g R}

Given

\frac{R_{A}}{R_{B}} = r

and

\frac{g_{A}}{g_{B}} = x

Taking ratio of the escape velocities of the 2 planets,

\frac{V_{A}}{V_{B}} = \sqrt{r x}

Formula

Escape Velocity in an Orbit

Consider a satellite of mass m, orbiting around the earth, then binding of the satellite on the surface of earth is given by

B . E . = \frac{G M m}{r}

r - radius of orbit
To escape from the orbit of earth, the Kinetic energy of satellite equivalent to binding energy.
K.E. = B.E.

\frac{1}{2} m {v_{e}}^{2} = \frac{G M m}{r}

v_{e} = \sqrt{\frac{2 G M}{r}}

..........(1)
We know that

g = \frac{G M}{r^{2}}

hence,

v_{e} = \sqrt{2 g r}

..........(2)

Example

Maximum Height Attained by a projectile

Example: A projectile is fired vertically upwards from the surface of the earth with a velocity

K V_{e}

, where

V_{e}

is the escape velocity and

K < 1

. If

R

is the radius of the earth, the maximum height to which it will rise measured from the centre of the earth will be (neglect air resistance)?

Solution:
According to the law of conservation of energy

\frac{1}{2} m V^{2} = \frac{m g h}{1 + \frac{h}{R}}

\frac{1}{2} m (K V_{e})^{2} = \frac{m g (r - R)}{1 + \frac{(r - R)}{R}}

\frac{1}{2} m K^{2} (2 g R) = \frac{m g (r - R)}{1 + \frac{(r - R)}{R}}

K^{2} R [1 + \frac{(r - R)}{R}] = r - R

K^{2} r = r - R

R = (1 - K^{2}) r

r = \frac{R}{1 - K^{2}}

Example

Escape Velocity for an object projected at an angle

Example: The escape velocity for a body projected vertically upwards from the surface of earth is

11

km/s. If the body is projected at an angle of

45^{0}

with the vertical, what will be the escape velocity?

Solution:Escape speed of a body from Earth's surface is given by

v_{m i n} = \sqrt{2 g R}

This expression is obtained by conservation of energy and doesn't involve in which direction the body is thrown/projected.
So, irrespective of the angle of projection, escape speed of the body from
Earth's surface remains constant i.e.

\approx 11

km/s

Example

Escape velocity of a body under force from two planets

Example:
If 'd' is the distance between the centers of the earth of mass

M_{1}

and moon of mass

M_{2}

, then find the velocity with which a body should be projected from the mid point of the line joining the earth and the moon, so that it just escapes.Solution:
Using law of conservation of energy
Kinetic energy minimum

=

- Potential energy
P.E at mid point

= \frac{- 2 G M m}{d} (M_{1} + M_{2})

= \frac{1}{2} m V_{e}^{2} = \frac{2 G m}{d} (M_{1} + M_{2})

\Rightarrow V_{e} = \sqrt{\frac{4 G (M_{1} + M_{2})}{d}}

Definition

Critical velocity

The horizontal velocity required to project a satellite to a height above the earth's surface, such that it attains a circular orbit around the earth is called the orbital velocity or critical velocity (