Question

The potential energy of a rocket of mass $100kg$ at height ${10}^{7}m$ from earth surface is $4\times {10}^{9}Joule$. The weight of the rocket at height ${10}^9$ will be

• A. $4\times {10}^{-2}N$
• B. $4\times {10}^{-3}N$
• C. $8\times {10}^{-2}N$
• D. $8\times {10}^{-3}N$

Answer 1

The correct option is A $4\times {10}^{-2}N$

Distance $R=r+h=0.64\times {10}^7+{10}^7=1.64\times {10}^{7}m$
Now$,U=\frac{GMm}{R}=\frac{GM\times 100}{1.64\times {10}^7}=GM\times .609\times {10}^{-5}$
But, given $U=4\times {10}^9\Rightarrow GM\times 0.609\times {10}^{-5}=4\times {10}^9$
$GM=6.57\times {10}^{14}$
now at height ${10}^{9}m$, $R=r+h=100.64\times {10}^7$
weight $mg=\frac{GMm}{R^2}=\frac{6.57\times {10}^{14}}{(100.64\times {10}^7{)}^2}=4\times {10}^{-2}$ N.