Question

Can we derive Acceleration due to Gravity. If yes then How?

Answer 1

Answer:

Explanation:

We know from Newton's second law that

$F=ma$

Now, Newton's law of universal gravitation says

$F=G\frac{mM_{}}{r^2}$

where m = mass of object 1, M = mass of object 2, and r = distance between the objects. '

Now, equating the two equations above gives

$ma=G\frac{mM}{r^2}$

Now, dividing both sides by m gives

$\boxed{a=G\frac{M}{r^2}}$

This is the value of the acceleration due to gravity due to an object of mass M at a distance r away from another object.

Now, near earth, the above equation gives us a = g = 9.8 m/s^2. This can be obtained by substituting M = 5.97 * 10^24 kg, r = earth radius = 6378 km, and G = 6.67 * 10^ -11.

$a=(6.67\cdot10^{-11})\cdot\frac{5.97\cdot10^{24}}{(6371\cdot1000)^2}$

$\Rightarrow a=\frac{6.67\cdot10^{-11}\cdot5.97\cdot10^{24}}{(6371\cdot10^3)^2}$

$\Rightarrow a=\frac{6.67\cdot5.97\cdot\cdot10^{-11}\cdot10^{-11}\cdot10^{24}}{6371^2\cdot10^6^{}}$

$\Rightarrow(\frac{6.67\cdot5.97}{6371^2})\cdot\frac{10^{-11}\cdot10^{24}}{10^6}$

$\Rightarrow a=\frac{6.67\cdot5.97}{(6.371\cdot10^3)^2}\cdot\frac{10^{-11}\cdot10^{24}}{10^6}$

$\Rightarrow a=\frac{6.67\cdot5.97}{6.371^2\cdot10^6}\cdot\frac{10^{-11}\cdot10^{24}}{10^6}$

$\Rightarrow a=\frac{6.67\cdot5.97}{6.371^2}\cdot\frac{10^{-11}\cdot10^{24}}{10^6\cdot10^6}$

$a\approx9.81m/s^2$

which when rounded to the nearest tenth gives

$\boxed{a=9.8m/s^2\text{.}}$