Question

what will be the acceleration due to gravity if mass of the object is double and radius of object is 3/4th​

Answer 1

Answer:

$a=\frac{G32m}{9r^2} ... 0r ... a= 3.5\frac{Gm}{r^2}$

Explanation:

$a=\frac{Gm}{r^2}$

$a=\frac{G 2m}{3/4r^2 }$

$a=\frac{G2m16}{9r^2}$

$a=\frac{G32m}{9r^2} ... 0r ... a= 3.5\frac{Gm}{r^2}$

Answer 2

Answer:

(32/9) g

Explanation:

g (acceleration due to gravity) is given by the formula

g = GM/ R²

where M is mass of the object whose g is to be calculated and R is the radius of the object (for finding g at the surface of the object)

hence,

new acceleration due to gravity (g') = G(2M)/ (3/4 R)²

= 32GM/9R²

= (32/9) g. (as g is GM/R²)

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