Question

weight of a body of mass m is decreased by 1% when it is raised to height h above the earth surface. if the body is taken to a depth h in a mine,then its weight will be??

Answer 1

sry its g not G

Answer 2

ans is 0.5% change

Answer 3

it is said it is 0.5%decreases i think u didnt read ques again by concn. as here it is said abu weight plz read it and explain me

Answer 4

The weight of a body of mass m in the gravitational field of the earth is
\[ W = \frac{GMm}{r^2} \]
where M is the mass of the earth and r is the radial distance of the object from the center of the earth

If the mass is at a height h over the surface of the earth, then its distance from the center of the earth is
r = R + h
where R is the radius of the earth.

The weight of an object on the earth is
\[ W = \frac{GMm}{R^2} \]

If at height h, i.e., r = R + h, it is 1% less, that means
\[ W = \frac{GMm}{r^2} = 0.99 \frac{GMm}{R^2} \]

Re-arranging this equation, this means that
\[ r^2 = (R+h)^2 = (1/0.99) R^2 \]
i.e.,
\[ R + h = \sqrt{1/0.99} R \]
i.e.,
\[ h = (\sqrt{1/0.99} - 1)R -- (*)\]

Now, if you're in a mine at a depth of h, the radial distance becomes
\[ r = R - h \]
Using the expression (*) for h,
\[ r = R - h = (2 - \sqrt{1/0.99})R \]
Calculating this, to five decimal places
\[ r = 0.99496R \]
Hence
\[ r^2 = 0.9899 R^2 \] and
\[ 1/r^2 = 1.01016 1/R^2 \]

and at this depth of r = R - h, the weight of the mass m is
\[ \frac{GMm}{r^2} = 1.01016 \frac{GMm}{r^2} \]

In other words, the mass is 1.016% more.

Answer 5

Correction of last equation:
\[\frac{GMm}{r^2} = 1.01016 \frac{GMm}{R^2} \]

Answer 6

:( the ans is 0.5% decreases from its intital wieght

Answer 7

thanX for afforts but i ll ask my teacher now i dont think i can get ans.. but thnx for ur help :)