Question

The value of acceleration due to gravity at a point P inside the earth and at another point Q outside the earth is g/2 .(g being acceleration due to gravity at the surface of the earth). Maximum possible distance in terms of radius of earth R between P and Q is:

• A. $2R$
• B. $2R(\sqrt{2}+1)$
• C. $\frac{R}{2}(2\sqrt{2}-1)$
• D. $\frac{R}{2}(2\sqrt{2}+1)$

Answer 1

The correct option is A $2R$

In above the surface,

$\frac{g}{2}=\frac{GM}{(R+h{)}^2}....1$

And for below the surface,

$\frac{g}{2}=\frac{GM}{(R-d{)}^2}....2$

Equating both the above equation then we get,

$R^2=h^2+2Rh$

And,

$R^2=d^2–2Rd$

By solving quadratic for both hand d terms as:

$h=(\sqrt{2}–1)R$

And,

$d=(\sqrt{2}+1)R$

Thus,

$h+d=(\sqrt{2}+1)R+(\sqrt{2}-1)R$

$=>2\sqrt{2}R$ meter.