Question
Explain how Newton concluded that gravitational force $F \propto=\frac{M m}{r^2}$
✓
Answer
Before generalising and stating universal law of gravitation, Newton first studied the motion of moon around the Earth.
$1.$The known facts about the moon were,
the time period of revolution of moon around the Earth $(T) = 27.3$ days.
distance between the Earth and the moon $(r) = 3.85 \times 10^5 km.$
the moon revolves around the Earth in almost circular orbit with constant angular velocity $ω.$
$2.$Thus, the centripetal force experienced by moon $($directed towards the centre of the Earth$)$ is given by,
$F = mrω^{2 }……… (1)$
Where, $m =$ mass of the moon
$3.$ From Newton’s laws of motion,
$F = ma$
$\therefore a = rω^{2 }………… (2)$
$4.$ But, $\omega=\frac{2 \pi}{T}$
$\therefore a=r\left(\frac{2 \pi}{T}\right)^2$
Substituting values of $r$ and $T$ , we get
$a =\frac{\left(3.85 \times 10^5 \times 10^3 \times 4 \pi^2\right)}{(27.3 \times 24 \times 60 \times 60)^2}$
$\therefore a=0.0027 \ m / s ^2$
$5.$This is the acceleration of the moon directed towards the centre of the Earth.
$6.$This acceleration is much smaller than the acceleration felt by bodies near the surface of the Earth $($while falling on Earth$).$
$7.$The value of acceleration due to Earth’s gravity at the surface is $9.8 \ m/s^2.$
$\therefore \frac{a_{\text {cbjact }}}{a_{\text {moon }}}=\frac{9.8}{0.0027} \approx 3600\ \ \ \ ......(3)$
Also,
$\frac{\text { distance of moon from the Earth's centre }}{\text { distance of object from the Earth's centre }}$
$=\frac{3.85 \times 10^5 \ km}{6378 \ km} \approx 60\ \ \ \ \ \ .......(4)$
Thus, from equations $(3)$ and $(4)$ we get
$\frac{ a _{\text {object }}}{ a _{\text {moon }}}=\left(\frac{\text { distance of moon }}{\text { distance of object }}\right)^2\ \ \ \ ......(5)$
$8.$Newton therefore concluded that the acceleration of an object towards the Earth is inversely proportional to the square of distance of object from the centre of the Earth.
$\therefore a \propto \frac{1}{r^2}$
As, $F = ma$
Therefore, the force exerted by the Earth on an object of mass $m$ at a distance $r$ from it is
$F \propto \frac{ m }{ r ^2}$
Similarly, an object also exerts a force on the Earth which is
$F_E \propto \frac{M}{r^2}$
Where $M$ is the mass of the Earth. .
$9.$According to Newton’s third law of motion, $F = F_E.$ Thus, $F$ is also proportional to the mass of the Earth. From these observations, Newton concluded that the gravitational force between the Earth and an object of mass $m$ is $F$ $\propto \frac{ Mm }{ r ^2}$
$1.$The known facts about the moon were,
the time period of revolution of moon around the Earth $(T) = 27.3$ days.
distance between the Earth and the moon $(r) = 3.85 \times 10^5 km.$
the moon revolves around the Earth in almost circular orbit with constant angular velocity $ω.$
$2.$Thus, the centripetal force experienced by moon $($directed towards the centre of the Earth$)$ is given by,
$F = mrω^{2 }……… (1)$
Where, $m =$ mass of the moon
$3.$ From Newton’s laws of motion,
$F = ma$
$\therefore a = rω^{2 }………… (2)$
$4.$ But, $\omega=\frac{2 \pi}{T}$
$\therefore a=r\left(\frac{2 \pi}{T}\right)^2$
Substituting values of $r$ and $T$ , we get
$a =\frac{\left(3.85 \times 10^5 \times 10^3 \times 4 \pi^2\right)}{(27.3 \times 24 \times 60 \times 60)^2}$
$\therefore a=0.0027 \ m / s ^2$
$5.$This is the acceleration of the moon directed towards the centre of the Earth.
$6.$This acceleration is much smaller than the acceleration felt by bodies near the surface of the Earth $($while falling on Earth$).$
$7.$The value of acceleration due to Earth’s gravity at the surface is $9.8 \ m/s^2.$
$\therefore \frac{a_{\text {cbjact }}}{a_{\text {moon }}}=\frac{9.8}{0.0027} \approx 3600\ \ \ \ ......(3)$
Also,
$\frac{\text { distance of moon from the Earth's centre }}{\text { distance of object from the Earth's centre }}$
$=\frac{3.85 \times 10^5 \ km}{6378 \ km} \approx 60\ \ \ \ \ \ .......(4)$
Thus, from equations $(3)$ and $(4)$ we get
$\frac{ a _{\text {object }}}{ a _{\text {moon }}}=\left(\frac{\text { distance of moon }}{\text { distance of object }}\right)^2\ \ \ \ ......(5)$
$8.$Newton therefore concluded that the acceleration of an object towards the Earth is inversely proportional to the square of distance of object from the centre of the Earth.
$\therefore a \propto \frac{1}{r^2}$
As, $F = ma$
Therefore, the force exerted by the Earth on an object of mass $m$ at a distance $r$ from it is
$F \propto \frac{ m }{ r ^2}$
Similarly, an object also exerts a force on the Earth which is
$F_E \propto \frac{M}{r^2}$
Where $M$ is the mass of the Earth. .
$9.$According to Newton’s third law of motion, $F = F_E.$ Thus, $F$ is also proportional to the mass of the Earth. From these observations, Newton concluded that the gravitational force between the Earth and an object of mass $m$ is $F$ $\propto \frac{ Mm }{ r ^2}$
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