Question:

The weight of an object at earth's surface is $700\, g\, wt$. What will be its weight at the surface of a planet whose radius is $1 / 2$ and mass is $1 / 7$ of that of the earth ?

Updated On: Jul 2, 2022
  • 200 g wt.
  • 400 g wt.
  • 50 g wt.
  • 300 g wt.
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The Correct Option is B

Solution and Explanation

Weight at earth's surface $w=m g$ Acceleration due to gravity at earth's surface $g=\frac{G M_{c}}{R_{e}^{2}}$ For planet, $g'=\frac{G M}{R^{2}}=\frac{G M_{c} / 7}{\left(R_{e} / 2\right)^{2}}$ $=\frac{4}{7} \frac{G m_{e}}{R_{e}^{2}}=\frac{4}{7} g$ $\therefore$ Weight at the surface of planet $=m g'$ $=m\left(\frac{4 g}{7}\right)$ $=\frac{4}{7} m g=\frac{4}{7} \times 700$ $=400\, g\, wt$
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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].