Question:
The value of $g$ at a height equal to half the radius of the earth from the earth's surface is
The value of $g$ at a height equal to half the radius of the earth from the earth's surface is
Updated On: Jun 21, 2022
- $\frac{g}{2}$
- $\frac{g}{3}$
- $\frac{4g}{9}$
- $\frac{g}{4}$
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The Correct Option is C
Solution and Explanation
Acceleration due to gravity at a height $h$
from the earth's surface is $g_{h} = \frac{gR^{2}_{E} }{\left(R_{E} +h\right)^{2}}$
where g is acceleration due to gravity on the earth's surface.
At $ h =\frac{R_{E}}{2} ,g_{h} = \frac{gR^{2}_{E}}{\left( R_{E} +\frac{R_{E}}{2}\right)^{2}} = \frac{4}{9}g$
from the earth's surface is $g_{h} = \frac{gR^{2}_{E} }{\left(R_{E} +h\right)^{2}}$
where g is acceleration due to gravity on the earth's surface.
At $ h =\frac{R_{E}}{2} ,g_{h} = \frac{gR^{2}_{E}}{\left( R_{E} +\frac{R_{E}}{2}\right)^{2}} = \frac{4}{9}g$
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Concepts Used:
Gravitation
In mechanics, the universal force of attraction acting between all matter is known as Gravity, also called gravitation, . It is the weakest known force in nature.
Newton’s Law of Gravitation
According to Newton’s law of gravitation, “Every particle in the universe attracts every other particle with a force whose magnitude is,
- F ∝ (M1M2) . . . . (1)
- (F ∝ 1/r2) . . . . (2)
On combining equations (1) and (2) we get,
F ∝ M1M2/r2
F = G × [M1M2]/r2 . . . . (7)
Or, f(r) = GM1M2/r2
The dimension formula of G is [M-1L3T-2].