Question:
Consider two solid spheres of radii $R, = 1m, \,R_2 = 2m$ and masses $M_1$ and $M_2$, respectively. The gravitational field due to sphere (1) and (2) are shown. The value of $\frac{m_{1}}{m_{2}}$ is :
Consider two solid spheres of radii $R, = 1m, \,R_2 = 2m$ and masses $M_1$ and $M_2$, respectively. The gravitational field due to sphere (1) and (2) are shown. The value of $\frac{m_{1}}{m_{2}}$ is :
Updated On: Jul 8, 2024
- $\frac{2}{3}$
- $\frac{1}{6}$
- $\frac{1}{2}$
- $\frac{1}{3}$
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The Correct Option is B
Solution and Explanation
Gravitational field on the surface of a solid
sphere $I_{g}=\frac{GM}{R^{2}}$
By the graph
$\frac{GM_{1}}{\left(1\right)^{2}}=2$
and $\frac{GM_{2}}{\left(2\right)^{2}}=3$
On solving
$\frac{M_{1}}{M_{2}}=\frac{1}{6}$
sphere $I_{g}=\frac{GM}{R^{2}}$
By the graph
$\frac{GM_{1}}{\left(1\right)^{2}}=2$
and $\frac{GM_{2}}{\left(2\right)^{2}}=3$
On solving
$\frac{M_{1}}{M_{2}}=\frac{1}{6}$
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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].