Question

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

Answer 1

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(mM)/(r^2)`

Now, dividing both sides by m gives

`\boxed{a=G(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(5.97\cdot10^(24))/((6371\cdot1000)^2)`

`\Rightarrow a=(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((6.67\cdot5.97)/(6371^2))\cdot(10^(-11)\cdot10^(24))/(10^6)`

`\Rightarrow a=(6.67\cdot5.97)/((6.371\cdot10^3)^2)\cdot(10^(-11)\cdot10^(24))/(10^6)`

`\Rightarrow a=(6.67\cdot5.97)/(6.371^2\cdot10^6)\cdot(10^(-11)\cdot10^(24))/(10^6)`

`\Rightarrow a=(6.67\cdot5.97)/(6.371^2)\cdot(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{.}}`