Consider a body which weighs w newton at the surface of the earth. Its weight at a height equals to half the radius of the earth, will be

Question

Consider a body which weighs w newton at the surface of the earth. Its weight at a height equals to half the radius of the earth, will be (A) (w/2) (B) (2 w/3) (C) (4 w/9) (D) (8 w/27)

Tardigrade Admin · Accepted Answer

The value of acceleration due to gravity at a height

h

above the earth's surface is given by

Solution

g^{''} = \frac{g}{( 1 + \frac{h}{R} ) ^{2}}

where

R

is radius of earth.
When

h = \frac{R}{2}

g^{''} = \frac{g}{1 + \frac{h ^{2}}{R}} = \frac{4 g}{9}

Hence, weight

w^{''} = m g^{''} = \frac{4}{9} m g = \frac{4}{9} w

.

Q. Consider a body which weighs $w$ newton at the surface of the earth. Its weight at a height equals to half the radius of the earth, will be

$\frac{w}{2}$

$\frac{2 w}{3}$

$\frac{4 w}{9}$

$\frac{8 w}{27}$

Q. Consider a body which weighs w newton at the surface of the earth. Its weight at a height equals to half the radius of the earth, will be

2w​

32w​

94w​

278w​

The value of acceleration due to gravity at a height h above the earth's surface is given by g′′=(1+Rh​)2g​ where R is radius of earth. When h=2R​ g′′=1+Rh2​g​=94g​ Hence, weight w′′=mg′′=94​mg=94​w.

Q. Consider a body which weighs $w$ newton at the surface of the earth. Its weight at a height equals to half the radius of the earth, will be

$\frac{w}{2}$

$\frac{2 w}{3}$

$\frac{4 w}{9}$

$\frac{8 w}{27}$