Laws of Gravitation – Class 11 Physics

Laws of Gravitation – Class 11 Physics

Gravitation or gravity is the most common of the four fundamental forces, we are familiar with. This is the force that acts on each and every single body having finite mass. On earth, gravity provides the weight to all physical objects and the moon’s gravity causes the ocean tides. The motion of celestial bodies such as planets, stars and satellites are all governed by the laws of gravity. Recently the Indian Space Research Organization (ISRO) have used the gravitational force to propel their interplanetary space vehicle, the Mars orbiter mission – Mangalyaan.

Tycho Brahe (1546-1601) and Johannes Kepler (1571-1630) studied the planetary motion in great detail and Kepler formulated his findings between the year 1609 and 1619 in three laws of planetary motion. In astronomy, these laws of planetary motion describe the motion of planets around the Sun.

Kepler’s Law of Planetary Motion

  1. The orbit of a planet is an ellipse with the Sun at one of the two foci.
  2. A line segment joining a planet and the sun sweeps out equal areas during equal intervals of time.
  3. The square of the orbital period of a planet is proportional to the cube of the semi-major axis of its orbit.

Newton’s Law of Gravitation

Later on, Sir Isaac Newton came up with a very daring assumption and proposed that the laws of nature are the same for every celestial body. This idea was revolutionary and generalized the Kepler’s law of planetary motion. Newton further proposed that the gravitational force between two bodies with mass m1 and m2 is proportional to the product of their masses and is inversely proportional to the square of the distance between them.

 F\propto m_1m_2 and  F\propto \frac{1}{r^2}

We introduce proportionality constant G to write the equation,

    \[ \boxed{F=G\frac{m_1m_2}{r^2}} \]

The constant G is called the universal constant of gravitation and its value is found to be 6.67\times10^{-11} N-m2/kg2. Newton’s law of Gravitation is often known as the universal law of gravitation. Newton’s theory of gravitation was first proved by the British scientist Henry Cavendish in the year 1798.

Gravitational Potential Energy

Gravitational potential energy is the potential an object has to do work as a result of being located at a particular position within the gravitational field. Considering Newton’s law of gravitation, the work done due to a small displacement  dr is

 {dW= -\frac{Gm_1m_2}{r^2}dr}

Now, the increase in potential energy is defined as  dU=-dW. Thus considering infinity as the reference point and integrating between r and infinity, we find the expression for the Gravitational potential energy due to a two particle system.

 U(r)= -\frac{Gm_1m_2}{r}

Here r is the separation between the particles with masses  m_1 and m_2.

Gravitational Potential

Gravitational potential is the change in potential energy per unit mass, as the mass is brought from a reference point to a given point r within the influence of the gravitational field.

  1. Gravitational potential due to a point mass M at a distance r is  -\frac{GM}{r}
  2. Gravitational potential due to a uniform ring at a point on its axis is  -\frac{GM}{\sqrt{a^2+r^2}}
  3. Gravitational potential due to a uniform thin spherical shell
    • Out side of the shell is  -\frac{GM}{r}
    • Inside of the shell is  -\frac{GM}{a}
  4. Gravitational potential due to a uniform solid sphere
    • Out side of the sphere is  -\frac{GM}{r}
    • Inside of the sphere is  -\frac{GM}{2a^3}(3a^2-r^2)
    • At the center of the sphere is  -\frac{3GM}{2a}

Gravitational Field

We define the intensity of gravitational field \vec{E} at a point as the force per unit mass. Thus if \vec{F} be the force due to mass m then the intensity of gravitational field \vec{E} is

 \vec{E}= -\frac{\vec{F}}{m}

  1. Gravitational field due to a point mass is  -\frac{GM}{r^3}\vec{r}
  2. Gravitational field due to a uniform ring at a point on its axis is  -\frac{GM}{(a^2+r^2)^{3/2}}\vec{r}
  3. Gravitational field due to a uniform disc at a point on its axis is  -\frac{2GM}{a^2}[\frac{1}{r}-\frac{1}{\sqrt{a^2+r^2}}]\vec{r}
  4. Gravitational field due to a uniform thin spherical shell
    • Out side of the shell is  -\frac{GM}{r^3}\vec{r}
    • Inside of the shell is 0
  5. Gravitational field due to a uniform solid sphere
    • Outside of the sphere is  -\frac{GM}{r^3}\vec{r}
    • Inside of the sphere is  -\frac{GM}{a^3}\vec{r}

Variation in the Value of Gravitational Acceleration

Gravitational acceleration is given by

    \[  g=\frac{F}{m}\]

where F is the force exerted by the earth on an object of mass m. This force can be affected by several factors. In the following section we will discuss about the factors which can change gravitational acceleration.

  1. Height from the surface of the earth
    The gravitational force on a mass m at a height h from the surface of the earth is given by

        \[  F=\frac{GMm}{(R+h)^2}\]

    Here M is the mass of earth and R is the radius. Therefore we can easily calculate the gravitational acceleration as

        \[  g(outside)=\frac{F}{m}=\frac{GM}{(R+h)^2}\]

    Thus we can see that the gravitational acceleration decreases as we go up from the surface of the earth. Counterintuitively gravitational acceleration also decreases as one goes inside the earth. The force due to gravity inside the surface of the earth is

        \[  F=\frac{GMm}{R^3}(R-h)\]

    or

        \[  g(inside)=\frac{GM}{R^2}\left(\frac{R-h}{R} \right)\]

    Therefore it becomes clear that the maximum value of g is at the surface of the earth and it decreases as we increase the height or depth.

  2. Rotational motion of the earth
    Centrifugal force associated with any rotational motion is given by m\omega^2 r. Here \omega is the angular velocity of the earth which is obviously constant for any object on earth and r is the radius of the circle in which the particle rotates. It is clear that r is maximum for any equatorial object and is minimum for any polar object. Therefore the effect of centrifugal force is maximum for any equatorial object and is minimum for a polar object. Thus resultant gravitational acceleration is minimum for an equatorial object and maximum for a polar object.
  3. Nonsphericity of the earth
    The earth is not a perfect sphere. The shape slightly deviates from a perfect sphere with an equatorial radius larger than the polar radius. The value of g is accordingly larger at poles.
  4. Nonuniformity of the earth
    The density of the earth is also not uniform and varies with the local concentrations of minerals, the position of water bodies etc. The local value of g also varies due to these nonuniformities.

Escape Velocity

Our everyday experiences tell us that when we throw a particle up, it goes up to a certain distance and returns back. However, it is possible to throw a particle with a certain velocity that it never returns back. The minimum initial velocity which is required is called the escape velocity. The escape velocity for any celestial body can be calculated very easily using the concept of conservation of total energy.

Here we will calculate the escape velocity for the earth. Let the initial velocity of the particle be u and the mass is m. Consequently, the corresponding kinetic energy of that particle is \frac{1}{2}mv^2. The gravitational potential energy at the surface of the earth is -\frac{GMm}{R}. Here M is the mass of the earth and R is the radius. At any point of the motion, the sum of the kinetic and the potential energy remains constant. Thus if the velocity of the particle at a certain height h from the surface of the earth be v. From the conservation of total energy

    \[  \frac{1}{2}mu^2 - \frac{GMm}{R}=\frac{1}{2}mv^2 - \frac{GMm}{R+h} \]

If the particle attains a certain maximum height from the surface of the earth then the velocity will be zero at this definite height. This maximum height can be calculated by equating the left-hand side of the above equation to zero. However, if we can make this value to be greater than zero, the velocity of the particle will never reach zero and it will continue to go farther away from the surface of the earth. Thus the particle will never return to the earth if

    \[\frac{1}{2}mu^2-\frac{GMm}{R} \ge 0\]

or,

    \[u\ge \sqrt{\frac{2GM}{R}} \]

Thus the critical velocity known as the escape velocity is

    \[u= \sqrt{\frac{2GM}{R}}\]

Cover image courtesy: Jet Propulsion Laboratory, NASA.

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