An object is taken to a height above the surface of earth at a distance (5/4) R from the centre of the earth. Where radius of earth, R =6400 km. The percentage decrease in the weight of the object will be

Question

An object is taken to a height above the surface of earth at a distance (5/4) R from the centre of the earth. Where radius of earth, R =6400 km. The percentage decrease in the weight of the object will be (A) 36 % (B) 50 % (C) 64 % (D) 25 %

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/f72f454810548c2d6324c0d6f00b6bea-.png" /> g text eff =( g /(1+( h ) R )2) ; g text eff =( g /(1+(1)4)2)=(16 g /25) change =( g text eff - g / g ) × 100=((16/25)-1/1) × 100 =(-9/25) × 100=-36 %

Q. An object is taken to a height above the surface of earth at a distance $\frac{5}{4} R$ from the centre of the earth. Where radius of earth, $R = 6400 km$ . The percentage decrease in the weight of the object will be

$36%$

$50%$

$64%$

$25%$

$g_{eff} = \frac{g}{( 1 + \frac{h}{R} ) ^{2}}; g_{eff} = \frac{g}{( 1 + \frac{1}{4} ) ^{2}} = \frac{16 g}{25}$
change $= \frac{g _{eff} - g}{g} \times 100 = \frac{\frac{16}{25} - 1}{1} \times 100$
$= \frac{- 9}{25} \times 100 = - 36%$

Q. An object is taken to a height above the surface of earth at a distance 45​R from the centre of the earth. Where radius of earth, R=6400km. The percentage decrease in the weight of the object will be

36%

50%

64%

25%