Question

Assuming the earth to be a sphere of uniform mass density. how much would it a body weight 800km below the surface of earth. if it's weight is 360 N on the surface given R= 6400 km.​

Answer 1

Question:-

Assuming the earth to be a sphere of uniform mass density. how much would it a body weight 800km below the surface of earth. if it's weight is 360 N on the surface given R= 6400 km.

Formula:- ${g \tiny{f}} = (1 - \frac{depth}{radius})g$

Answer:-

We know:-

1. g = 10 m/s² on earth surface
2. R=6400 (Radius of earth)(given)
3. Weight on earth surface= 360N (given)
4. depth = 800 (given)

Solution:-

${g \tiny{f}} = (1 - \frac{depth}{radius})g$

Putting the value of depth and Radius and g in formula .

${g \tiny{f}} = (1 - \frac{800}{6400})10 \\{g \tiny{f}} = (1 - \frac{ \cancel{800}^{ \: \: 1} }{{ \cancel{6400}}^{ \: \: 2}} )10 \\ {g \tiny{f}} = (1 - \frac{1}{2} )10 \\ {g \tiny{f}} =( \frac{2 - 1}{2} )10 \\ {g \tiny{f}} = ( \frac{1}{2} ) \times 10 \\ {g \tiny{f}} = 5 \: m {s}^{ - 2}$

For mass :-

$weight = m \times g \: (on \: surface) \\ 360 = m \times 10 \\ 36 = m$

For weight at depth of 800 Km

$weight \: = m \times g \\ w = 36 \times 5 \\ w = 180 \: newton \: (N)$

Answer 2

Answer:

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Explanation:

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