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BITSAT 2016
Exam
BITSAT
Year
2016
Subject
Physics
Chapter
Gravitation

BITSAT 2016 · Physics · Gravitation

Kepler's third law states that square of period of revolution (T)...

Kepler's third law states that square of period of revolution (T) of a planet around the sun, is proportional to third power of average distance r\mathrm{r} between sun and planet i.e. T2=Kr3\mathrm{T}^{2}=\mathrm{Kr}^{3} here K\mathrm{K} is constant. If the masses of sun and planet are M\mathrm{M} and m\mathrm{m} respectively then as per Newton's law of gravitation force of attraction between them is F=GMm\mathrmr2\mathrm{F}=\frac{\mathrm{GMm}}{\mathrmr^{2}}, here G\mathrm{G} is gravitational constant. The relation between G\mathrm{G} and K\mathrm{K} is described as

Options

  1. AGMK=4π2\mathrm{GMK}=4 \pi^{2}
  2. BK=G\quad \mathrm{K}=\mathrm{G}
  3. CK=1G\mathrm{K}=\frac{1}{\mathrm{G}}
  4. DGK=4π2\quad \mathrm{GK}=4 \pi^{2}

Correct answer

A. GMK=4π2\mathrm{GMK}=4 \pi^{2}

Step-by-step solution

As we know, orbital speed, Vorb = GM  r V_{\text{orb }}=\sqrt{\frac{\text{ GM }}{\text{ r }}} Time period T=2πrvorb =2πrGMr\mathrm{T}=\frac{2 \pi \mathrm{r}}{\mathrm{v}_{\text{orb }}}=\frac{2 \pi \mathrm{r}}{\sqrt{\mathrm{GM}}} \sqrt{\mathrm{r}} Squarring both sides, T2=(2πrrGM)2=4π2GM⋅r3⇒T2r3=4π2GM=K⇒GMK=4π2\mathrm{T}^{2}=\left(\frac{2 \pi \mathrm{r} \sqrt{\mathrm{r}}}{\sqrt{\mathrm{GM}}}\right)^{2}=\frac{4 \pi^{2}}{\mathrm{GM}} \cdot \mathrm{r}^{3}\Rightarrow \quad \frac{\mathrm{T}^{2}}{\mathrm{r}^{3}}=\frac{4 \pi^{2}}{\mathrm{GM}}=\mathrm{K} \Rightarrow \mathrm{GMK}=4 \pi^{2}
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