Where is the acceleration due to gravity minimum and maximum on the surface of the earth?

Open in App

Solution

Acceleration due to gravity is given by $g = \frac{G M}{R^{2}}$ where G is the universal gravitational constant, M is the mass of the earth and R is the radius of the earth.
Acceleration due to gravity is inversely proportional to the square of the distance from the centre of the earth.
More the distance from the centre of the earth lesser will be the value of acceleration due to gravity or vice versa.
The radius at the equator is more than that of the poles.
More the radius lesser is the value of acceleration due to gravity therefore acceleration due to gravity at the equator is less than the acceleration due to gravity at the poles.
Hence, the acceleration due to gravity is minimum at the equator and maximum at the poles on the surface of the earth

Suggest Corrections

Similar questions

9 . S H o w t h a t t h e w e i g h t o f a n o b j e c t o n t h e m o o n i s \frac{1}{6} t h o f i t s w e i g h t o n t h e e a r t h . 12 . W h e r e i s t h e a c c e l e r a t i o n d u e t o g r a v i t y m i n i m u m a n d m a x i m u m o n t h e s u r f a c e o f t h e e a r t h ?

\frac{g}{4}

g

R

Join BYJU'S Learning Program