It states that everybody in the universe attracts every other body with a force which id directly proportional to the product of their masses and inversely proportional to the square of the distance between their centres. Consider two bodies of bodies of masses m_{1} and m_{2} at a distance of r between them as shown in the figure.
According to the Newton’s law of gravitation,
The force between them is
\begin{align*} F &\propto m_1\times m_2 \dots (i) \\ F &\propto \frac {1}{r^2} \dots (ii) \\ \text {Combining equations} (i) \text {and} (ii), \text {we get} \\ F &\propto \frac {m_1m_2}{r^2} \\ \text {or,}\:F &= G\frac {m_1m_2}{r^2} \\ \end{align*}
where G is a constant of proportionality called the universal gravitational constant.
If m_{1} = m_{2} = 1 and r=1, then
$$F=G$$
Thus, universal gravitational constant, G equivalent to the force of attraction between two bodies each of mass unity and a unit distance apart. Its value is 6.67 ×10^{-11} Nm^{2} kg^{-2}, which indicates that gravitational force is very weak.
The gravitational force is independent of the intervening medium between the bodies. It follows the Newton’s third law of motion. For example, when a body A exerts a force on another body B, the body B also exerts the force A of the same magnitude but in opposite direction.
Gravitation and Gravity
Gravitation is the force of attraction between any two bodies in the universe where as gravity is the force due to earth on a body lying on or near the surface of the earth.
Acceleration due to Gravity
If an object is left from a height on the earth, it will accelerate towards the earth’s surface produces acceleration on it. The acceleration produced on an object due to earth’s gravity is called acceleration due to gravity and is denoted by g.
Relation between g and G
Consider an object of mass n lying on the surface of the earth as in the figure. Let M be the mass of the earth and R its radius. We can suppose that the earth is a sphere and its mass of concentrated at its centre. From Newton’s law of gravitation, the force of attraction between the earth and object is
\begin{align*} F&= \frac {GMm}{R^2} \\\end{align*}
If the g is the acceleration produced by this force, then \begin{align*} g &= \frac Fm \\ \text {or}\: g &= \frac {GMm}{R^2m} \\ \text {or,}\: g &= \frac {GM}{R^2} \\ \end{align*}
This relation shows that acceleration due to gravity depends only on the mass and radius of the earth and is independent of the mass of the falling object. That is why bodies with whatever their masses, will fall with the same acceleration due to gravity g near the earth’s surface.
\begin{align*} \text {As we know} \\ g &= \frac {GM}{R^2} \\ \text {or,} \: M &= \frac {gR^2}{G} \\ \text {We know that,} \\ \text {Radius of the earth, R} &= 6.4 \times 10^6 m \\ \text {Gravitational constant, G} &= 6.67\times 10^{-11} Nm^2Kg^{-2} \\ \text {Acceleration due to gravity, g} &= 9.8ms^{-2} \\ \\\end{align*} Then, mass of the earth \begin{align*} \\ M &= \frac {gR^2}{G} = \frac {9.8 \times (6.4 \times 10^6)^2}{6.67 \times 10^{-11}} = 6.018 \times 10^{24} kg \\ \\\end{align*}
If we consider the earth to be a spherical body of radius R and of uniform density\begin{align*}\: \rho , \text {then} \\ \rho &= \frac {mass}{volume} = \frac {M}{4/3\pi R^3} =\frac {3M}{4\pi R^3} = \frac {3g}{4\pi RG} \\ \rho &= \frac {3g}{4\pi RG} = \frac {3\times 9.8}{4\pi \times 6.4 \times 10^6 \times 6.67 \times 10^{-11} } = 5.5\times 10^3 kg m^{-3} \\\end{align*}
\begin{align*} F &\propto m_1\times m_2 \dots (i) \\ F &\propto \frac {1}{r^2} \dots (ii) \\ \text {Combining equations} (i) \text {and} (ii), \text {we get} \\ F &\propto \frac {m_1m_2}{r^2} \\ \text {or,}\:F &= G\frac {m_1m_2}{r^2} \\ \end{align*}
where G is a constant of proportionality called the universal gravitational constant.
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gravitational potantial
Mar 31, 2017
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