Question

If error in measuring the distance of a planet from sun is 0.4% and in measuring the mass of sun is 0.6% , then error in measuring the time period of planet around the sun is

• A. 1.8%
• B. 0.6%
• C. 1%
• D. 0.9%

Answer 1

Answer: (D)

0.9%

$\begin{array}{c}T=\text{Time period}\\ r=\text{radial distance between}\\ \text{sun and earth}\\ M=\text{mass of sum}\\ G=\text{Gravitational consteunt}\end{array}$

$\begin{array}{cc}\frac{\delta r}{r}\times \log \%& =0.4\%\\ \frac{\delta m}{m}\times 100& =0.6\%\end{array}$

$\begin{array}{c}T=\frac{2\pi }{\sqrt{Gm}}{r}^{3/2}\\ \text{Taking log both side}\\ \log T=\log \left[\frac{2\pi r^{3/2}}{(Gm{)}^{1/2}}\right]\end{array}$

$\begin{array}{c}\log T=\log 2+\log \pi +\frac{3}{2}\log r-\frac{1}{2}\log G-\frac{1}{2}\log m\\ \text{Differentiating both side:}\\ \frac{\delta T}{T}=0+0+\frac{3}{2}\frac{\delta r}{r}-0-\frac{1}{2}\frac{\delta m}{m}\end{array}$

$\begin{array}{c}\text{Error is always added:-}\\ \frac{\delta T}{T}=\frac{3}{2}\frac{\delta r}{r}+\frac{1}{2}\frac{\delta m}{m}\\ \frac{\delta T}{T}\times 100\%=\frac{3}{2}\frac{\delta r}{r}\times 100\%+\frac{1}{2}\frac{\delta m}{m}\times 100\%\end{array}$

$\begin{array}{cc}=& \frac{3}{2}\times 0.4+\frac{1}{2}\times 0.6\\ & =0.6+0.3\\ =& 0.9\%\text{.......Ans.}\end{array}$