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Mean solar day is the time interval between two successive noon when sun passes through zenith point (meridian). Sidereal day is the time interval between two successive transit of a distant star through the zenith point (meridian). By drawing appropriate diagram showing the earth's spin and orbital motion, show that mean solar day is $4\,\min$ longer than the sidereal day. In other words, distant stars would rise $4\,\min$ early every successive day
Solution
Every day the earth advances in the orbit by approximately $1^{\circ}$. Then, it will have to rotate by
$361^{\circ}$ (which we define as 1 day) to have the sun at zenith point again.
$\because \text { To cover } 361^{\circ} \text {, time taken }=24 \mathrm{~h}$ $\because \text { To cover } 1^{\circ} \text {, time taken }=t \text {, }$ $t=\frac{24}{361} \times 1=0.066 \mathrm{~h}$ $=3.99 \mathrm{~min}$ $\approx 4 \mathrm{~min}$
Hence, distant stars would rise $4 \mathrm{~min}$ early every successive day.
