Question:

Two satellites $A$ and $B$ go round a planet $P$ in circular orbits having radii $4R$ and $R$ respectively. If the speed of the satellite $A$ is $3\upsilon$, the speed of satellite $B$ will be

Updated On: Aug 15, 2022
  • $12 \upsilon$
  • $6 \upsilon$
  • $(4/3)\upsilon$
  • $(3/2)\upsilon$
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The Correct Option is B

Solution and Explanation

Orbit speed of the satellite around the earth is $\upsilon = \sqrt{ \frac{ GM}{ r}}$ where, G = Universal gravitational constant M = Mass of earth r = Radius of the orbit of the satellite For satellite A $ r_A = 4 R , v_A = 3V $ $ {\upsilon}_A = \sqrt{ \frac{GM}{ r_A}}$ ...(i) For satellite B $ r_B= R , v_ B =? $ $ {\upsilon}_B = \sqrt{ \frac{GM}{ r_B}}$ ...(ii) Dividing equation (ii) by equation (i), we get $\therefore \, \, \, \, \frac{ {\upsilon}_B}{ {\upsilon}_A}= \sqrt{\frac{r_A}{r_B}}$ or ${\upsilon}_B = {\upsilon}_A \sqrt{\frac{ r_A}{ r_B}}$ Substituting the given values, we get $\upsilon_B = 3 V \sqrt{\frac{4R}{ R}} \, \, \,$ or $\, \, \, \upsilon_B = 6V $
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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].