A body weighing $100 ext{ N}$ on the surface of the Earth weighs $x ext{ kg m s}^{-2}$ at a height $\frac{

Question

A body weighing $100 	ext{ N}$ on the surface of the Earth weighs $x 	ext{ kg m s}^{-2}$ at a height $\frac{1}{9}R_E$ above the surface of Earth. The value of $x$ is: (take $g=10 	ext{ m s}^{-2}$ at the surface of Earth and $R_E$ is the radius of Earth)

Accepted Answer

1. **Formula:** The acceleration due to gravity at a height $h$ above the surface of the Earth is given by $g(h) = \frac{g}{(1 + h/R_E)^2}$ .
2. **Weight Relationship:** The weight of a body is $W = mg$. Therefore, the weight at height $h$ is $W_h = m g(h) = \frac{mg}{(1 + h/R_E)^2} = \frac{W_{surface}}{(1 + h/R_E)^2}$.
3. **Substitution:** Given the weight on the surface $W_{surface} = 100 	ext{ N}$ and height $h = \frac{1}{9}R_E$.
   $$ W_h = \frac{100}{\left(1 + \frac{1/9 R_E}{R_E}ight)^2} = \frac{100}{\left(1 + \frac{1}{9}ight)^2} $$
4. **Calculation:**
   $$ W_h = \frac{100}{\left(\frac{10}{9}ight)^2} = \frac{100}{\frac{100}{81}} = 100 	imes \frac{81}{100} = 81 $$
   The weight is $81 	ext{ N}$ (or $81 	ext{ kg m s}^{-2}$). Thus, $x = 81$.

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