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${\displaystyle {\frac {1}{{\sqrt {18}}-{\sqrt {32}}}}}$

${\displaystyle ={\frac {1}{3{\sqrt {2}}-4{\sqrt {2}}}}}$

${\displaystyle ={\frac {-1}{\sqrt {2}}}}$

## Universal law of gravitation

The universal law of gravitation states that the force of gravitation is directly proportional to the product of masses and inversely proportional to the square of the distance between them.

### Mathematical Derivation of Universal Law of Gravitation

Let,

${\displaystyle {\mathit {f}}}$  = Gravitational force
${\displaystyle {\mathit {G}}}$  = Universal Gravitational Constant (${\displaystyle 6.67\times 10^{-11}Nm^{2}kg^{-2}}$ )
${\displaystyle {\mathit {M}},{\mathit {m}}}$  = Masses of the bodies
${\displaystyle {\mathit {d}}}$  = Seperation of the bodies.

therefore according to the law,

${\displaystyle {\mathit {f}}\propto {\frac {Mm}{d^{2}}}}$

${\displaystyle {\mathit {f}}=G{\frac {Mm}{d^{2}}}}$

### Gravity exerted by earth on moon

Mass of earth = ${\displaystyle M_{e}=6\times 10^{24}kg}$
Mass of moon = ${\displaystyle M_{m}=7.4\times 10^{22}kg}$
Seperation = ${\displaystyle 3.8\times 10^{8}m}$

According to the law,

${\displaystyle {\mathit {f}}=G{\frac {Mm}{d^{2}}}}$

${\displaystyle {\mathit {f}}=G{\frac {M_{e}M_{m}}{d^{2}}}}$

${\displaystyle {\mathit {f}}={\frac {6.67\times 10^{-11}Nm^{2}kg^{-2}\times 6\times 10^{24}kg\times 7.4\times 10^{22}kg}{3.8\times 10^{8}m\times 3.8\times 10^{8}m}}}$

${\displaystyle {\mathit {f}}=2.051\times 10^{20}}$

## Mathematical Numericals of sound

### A stone was dropped from a tower 500m high into a pond at the base of tower. When is the splash heard at top? take g = 10ms-2 and vs = 340ms-1

Time taken by stone to reach pond = t'

${\displaystyle s=ut+{\frac {1}{2}}at'^{2}}$

${\displaystyle 500m={\frac {10ms^{-2}\times t'^{2}}{2}}}$

${\displaystyle t'^{2}={\frac {500m\times 2}{10ms^{-2}}}}$

${\displaystyle t'^{2}={\frac {100}{s^{-2}}}=100s^{2}}$

${\displaystyle t'={\sqrt {100s^{2}}}=10s}$

Time taken by sound to reach the person = t

${\displaystyle v_{s}={\frac {s}{t''}}}$

${\displaystyle 340ms^{-1}={\frac {500m}{t''}}}$

${\displaystyle t''={\frac {500m}{340ms^{-1}}}}$

${\displaystyle t''=1.47s}$

Therefore total time taken

i.e. ${\displaystyle t=t'+t''=10s+1.47s=11.47s}$