Question

A body weight w newton at the surface of earth. Its weight at a height equal to half the radius of the earth will be

Answer 1

First we have to derive the relationship between the weight of the body and the radius of the earth.

We know that the weight of the body is equal to the gravitational force of attraction between that body and the centre of the earth. So we have,

$mg=\dfrac {GMm}{R^2}\\\\\\g=\dfrac {GM}{R^2}$

Since G and M are constants,

$g\propto\dfrac {1}{R^2}$

Thus the acceleration due to gravity is inversely proportional to the square of the radius of the earth. Then we can say that, when the radius becomes k times, then g becomes (1/k²) times.

Let k = 1 / 2. Then 1 / k² = 4.

Therefore, the acceleration due to gravity, g, becomes 4 times when the radius is halved.

Then the weight of the body will be, m(4g) = 4mg = 4w.

But this weight is obtained when the height is measured from the centre of the earth. What about if it is measured from the surface of the earth?

Here the height will be R + (R / 2) = 3R / 2.

I.e., the radius becomes 3 / 2 times. So let k = 3 / 2.

Then 1 / k² = (2 / 3)² = 4 / 9.

I.e., g becomes 4 / 9 times. Hence the weight will be,

m (4g / 9) = 4mg / 9 = 4w / 9.

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