Question 116: Gravitation

Question

The weight of a body on the surface of the earth is $$100 \mathrm{~N}$$. The gravitational force on it when taken at a height, from the surface of earth, equal to one-fourth the radius of the earth is:

Accepted Answer

To find the gravitational force on the body when taken at a height equal to one-fourth the radius of the Earth, we can use the formula for gravitational force:
$$F = G \frac{m_1 m_2}{r^2}$$
where $$F$$ is the gravitational force, $$G$$ is the gravitational constant, $$m_1$$ and $$m_2$$ are the masses of the two objects, and $$r$$ is the distance between their centers.
The weight of the body on the surface of the Earth is given as $$100\,	ext{N}$$, which is also the gravitational force acting on it:
$$F_	ext{surface} = G \frac{m_	ext{body} m_	ext{earth}}{R_	ext{earth}^2}$$
When the body is taken to a height equal to one-fourth the radius of the Earth, the distance between the centers of the body and the Earth becomes $$r_	ext{new} = R_	ext{earth} + \frac{1}{4} R_	ext{earth} = \frac{5}{4} R_	ext{earth}$$.
Now the gravitational force acting on the body at this new height is:
$$F_	ext{new} = G \frac{m_	ext{body} m_	ext{earth}}{r_	ext{new}^2}$$
$$F_	ext{new} = G \frac{m_	ext{body} m_	ext{earth}}{\left(\frac{5}{4} R_	ext{earth}ight)^2}$$
To find the ratio between the new gravitational force and the original force on the surface, we can write:
$$\frac{F_	ext{new}}{F_	ext{surface}} = \frac{G \frac{m_	ext{body} m_	ext{earth}}{\left(\frac{5}{4} R_	ext{earth}ight)^2}}{G \frac{m_	ext{body} m_	ext{earth}}{R_	ext{earth}^2}}$$
Canceling out the common terms, we get:
$$\frac{F_	ext{new}}{F_	ext{surface}} = \frac{R_	ext{earth}^2}{\left(\frac{5}{4} R_	ext{earth}ight)^2} = \frac{1}{\left(\frac{5}{4}ight)^2} = \frac{1}{\left(\frac{25}{16}ight)} = \frac{16}{25}$$
Now, since the weight of the body on the surface is $$100\,	ext{N}$$, we can find the new gravitational force as:
$$F_	ext{new} = \frac{16}{25} 	imes 100\,	ext{N} = 64\,	ext{N}$$
So, the gravitational force on the body when taken at a height equal to one-fourth the radius of the Earth is $$64\,	ext{N}$$.

Answer

50 N

Answer

64 N

Answer

25 N

Answer

100 N

Gravitation

The weight of a body on the surface of the earth is $100 \mathrm{~N}$. The gravitational force on it when taken at a height, from the surface of earth, equal to one-fourth the radius of the earth is:

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