Question

if the diameter of earth becomes doubles of its present value without change in it mean denisty, the new value of accleeration due to gravity on its surface will be ​

Answer 1

Answer:

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Explanation:

Okay we know that weight of an object = mg (Mass of the object x Acceleration due to gravity)

Okay now we know that accleration due to gravity (g) is given by:

g= GM/(R^2) (Let this be the first equation)

Where G is the gravitational constant

M is the mass of the Planet

R is the radius of the planet

Now changing the first equation we get

g= 4GM/(D^2)

where D is the diameter of the planet (R=D/2)

Now applying the conditions given in the question

Df (final diameter) = 2D

So g would be given as

g(f)= 4GM/(4D^2) (here gf is the final accleration due to gravity)

g(f)= GM/D^2

so comparing g(f) and g, we can clearly see that… g= 4 g(f)

so g(f)= g/4

That is we can say that the wight of the object would be 1/4 (one-fourth of its initial value).

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