Question

☆✿皿 Establish “g = GM//R²” 皿✿☆

Answer 1

To derive:

• g = GM / R²

Solution:

Suppose

• mass of an object placed on the surface of celestial body = m
• mass of celestial body = M
• it's radius (Distance between object and centre of celestial body) = R
• and, universal gravitational constant = G

then,

➡ Weight of the body = Gravitational force exerted by celestial body on the object

[So, Using Newton's second law and Newton's universal law of gravitation]

➡ mg = GMm/R²

➡ mg/m = GM / R²

➡ g = GM / R²

That is the expression for acceleration due to gravity on any planet or celestial body with mass M and radius R.

Answer 2

Answer:

Given :-

Derive :-

$\sf \: g = \dfrac{GM}{{R}^{2} }$

SoluTion :-

Assume :-

• Mass of any object placed on any object on celestial bodies as m
• Mass of celestial body = M
• Radius = R
• Gravitional constant of universe = G

Now,

Weight of Body = Gravitional force exerted by celestial bodies on object

mg = GMm/R²

mg/m = GM/R²

g = GM/R²