Question

A body weighs W newton on the surface of the earth. Its weight at a height equal to half the radius of the earth will be :

• A. $\frac{5W}{6}$
• B. $\frac{W}{6}$
• C. $\frac{3W}{2}$
• D. $\frac{4W}{9}$

Answer 1

Answer: (D)

Weight on earth is $\omega =mg=\frac{GmM}{R^2}$ ...(i) $\because$ $g=\frac{GM}{R^2}$ Now, at a height h above the earth surface, the value of g will become $g=\frac{GM}{{(R+h)}^2}$ Now, weight of the body at this height is given by $\omega =mg=\frac{GnM}{{(R+h)}^2}$ (ii) Now, dividing equation (ii) by (i) we obtam $\frac{\omega }{\omega }=\frac{R}{R+h}$ $h=\frac{R}{2}$ given, $\omega =\omega \frac{R^2}{{\left(R+\frac{R}{2}\right)}^2}=\frac{4\omega }{9}$