Question
State Newton's law of Gravitation. Find the percentage decrease in the weight of the body when taken to a height of 16km. above the surface of the earth. Radius of the earth is 6400km.
✓
Answer
It states that every body in the universe attracts every other body with a force which is directly propotional to the product of their masses and inversely propotional to square of distance between them.
$\text{F}=\frac{\text{Gm}_1\text{m}_2}{\text{r}^2}$
m1 and m2 → mass of two bodies
r → distance between two bodies.
The acceleration due to gravity at a height 'h' above the surface of the earth is
$\text{g}'=\text{g}\Big(1-\frac{2\text{h}}{\text{R}}\Big)$
$\text{g}-\text{g}'=\Big(\frac{2\text{hg}}{\text{R}}\Big)$
$\frac{\text{mg}-\text{mg}'}{\text{mg}}\times100=\frac{\text{g}-\text{g}'}{\text{g}}\times100$
$=\frac{2\text{h}}{\text{R}}\times100$
$=\frac{2\times16}{6400}\times100=0.5\%$
$\text{F}=\frac{\text{Gm}_1\text{m}_2}{\text{r}^2}$
m1 and m2 → mass of two bodies
r → distance between two bodies.
The acceleration due to gravity at a height 'h' above the surface of the earth is
$\text{g}'=\text{g}\Big(1-\frac{2\text{h}}{\text{R}}\Big)$
$\text{g}-\text{g}'=\Big(\frac{2\text{hg}}{\text{R}}\Big)$
$\frac{\text{mg}-\text{mg}'}{\text{mg}}\times100=\frac{\text{g}-\text{g}'}{\text{g}}\times100$
$=\frac{2\text{h}}{\text{R}}\times100$
$=\frac{2\times16}{6400}\times100=0.5\%$
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