A body weighs 72 N on surface of the earth. When it is taken to a height of h = 2R, where R is radius of earth, it would weigh

Question

A body weighs 72 N on surface of the earth. When it is taken to a height of h = 2R, where R is radius of earth, it would weigh (A) 36 N (B) 18 N (C) 9 N (D) 8 N

Tardigrade Admin · Accepted Answer

Weight on earth =m g=m × (G M/R2)=72 N Weight at height, h=2 R will be m g'=m((G M/r2))=m × (G M/(R+2 R)2) =(G M m/9 R2)=(72/9)=8 N,

Q. A body weighs $72 N$ on surface of the earth. When it is taken to a height of $h = 2 R$ , where $R$ is radius of earth, it would weigh

36 N

18 N

9 N

8 N

Weight on earth $= m g = m \times \frac{GM}{R ^{2}} = 72 N$
Weight at height, $h = 2 R$ will be $m g^{'} = m (\frac{GM}{r ^{2}}) = m \times \frac{GM}{( R + 2 R ) ^{2}}$
$= \frac{GM m}{9 R ^{2}} = \frac{72}{9} = 8 N$ ,

Q. A body weighs 72N on surface of the earth. When it is taken to a height of h=2R, where R is radius of earth, it would weigh

36 N

18 N

9 N

8 N

Weight on earth =mg=m×R2GM​=72N Weight at height, h=2R will be mg′=m(r2GM​)=m×(R+2R)2GM​ =9R2GMm​=972​=8N,

Q. A body weighs $72 N$ on surface of the earth. When it is taken to a height of $h = 2 R$ , where $R$ is radius of earth, it would weigh

Weight on earth $= m g = m \times \frac{GM}{R ^{2}} = 72 N$
Weight at height, $h = 2 R$ will be $m g^{'} = m (\frac{GM}{r ^{2}}) = m \times \frac{GM}{( R + 2 R ) ^{2}}$
$= \frac{GM m}{9 R ^{2}} = \frac{72}{9} = 8 N$ ,