Question 144: Gravitation

Question

An object is taken to a height above the surface of earth at a distance $${5 \over 4}$$ 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 :

Accepted Answer

The weight of an object at a distance d from the center of the Earth is given by:
$$W' = W \left(\frac{R}{d}ight)^2$$
where:
W' is the weight at distance d,
W is the weight at the Earth's surface,
R is the radius of the Earth,
d is the distance from the center of the Earth.
In this problem, we are asked to find the percentage decrease in weight, which can be calculated by:
$$\Delta W = \frac{W - W'}{W} 	imes 100\%$$
where ÃÂW is the percentage change in weight. Substituting the weight formula into the percentage change formula gives:
$$\Delta W = \left(1 - \left(\frac{R}{d}ight)^2ight) 	imes 100\%$$
The radius of the Earth R is 6400 km and the distance d from the center of the Earth is given as $\frac{5}{4}R$. Substituting these values into the equation gives:
$$\Delta W = \left(1 - \left(\frac{6400~km}{\frac{5}{4} \cdot 6400~km}ight)^2ight) 	imes 100\% = \left(1 - \left(\frac{4}{5}ight)^2ight) 	imes 100\% = 36\% $$
Therefore, the percentage decrease in the weight of the object when taken to a height of $\frac{1}{4}R$ above the surface of the Earth is 36%.

Answer

36%

Answer

50%

Answer

64%

Answer

25%

Gravitation

An object is taken to a height above the surface of earth at a distance ${5 \over 4}$ 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 :

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