Question

If a man weighs 90 N on the surface of the earth, then
the height above the surface of the earth of radius R.
where the weight is 30 N is​

Answer 1

Answer:

We know

$g = G \frac{ M}{ {r}^{2} }$

Now Weight , W = mg = 90

At surface g = 10 , so m = 90/10 = 9 kg

Now At height let value of gravitational acceleration is g'

Weight , 9g' = 30 so g' = 30/9 = 10/3

And g/g' = 3.

But

g is inversely proportional to distance from centre of earth

So

$\frac{g}{g} = \frac{ \frac{1}{ R {}^{2} } }{ \frac{1}{ (R + x) {}^{2} } } = \frac{3}{1} \\ \\ (R + x) {}^{2} = 3(R ) {}^{2} \\ \\ (R + x) {}^{2} - (R \sqrt{3} ) {}^{2} = 0 \\ \\ (x + R + R \sqrt{3} )(x + R - R \sqrt{3} ) = 0 \\ \\ since \: \: height \: \: cant \: \: be \: \: negative \\ \\ x = R( \sqrt{3} - 1)$

Answer 2

Explanation:

g'=g/2

g/2=g/(1+h/R)²

(1+h/R)²= 2

1+h/R=√2

h/R=√2-1

h/R=1.41-1=0.41

h=0.41R

Ans:(A) 0.41 R