Question.
When the planet Jupiter is at a distance of $824.7$ million kilometers from the Earth, its angular diameter is measured to be $35.72^{\prime \prime}$ of arc. Calculate the diameter of Jupiter.
When the planet Jupiter is at a distance of $824.7$ million kilometers from the Earth, its angular diameter is measured to be $35.72^{\prime \prime}$ of arc. Calculate the diameter of Jupiter.
solution:
Distance of Jupiter from the Earth, $D=824.7 \times 10^{6} \mathrm{~km}=824.7 \times 10^{9} \mathrm{~m}$
Angular diameter $=35.72^{\prime \prime}=35.72 \times 4.874 \times 10^{-6} \mathrm{rad}$
Diameter of Jupiter = d
Using the relation,
$\theta=\frac{d}{D}$
$d=\theta D=824.7 \times 10^{9} \times 35.72 \times 4.872 \times 10^{-6}$
$=143520.76 \times 10^{3}=1.435 \times 10^{5} \mathrm{~km}$
Distance of Jupiter from the Earth, $D=824.7 \times 10^{6} \mathrm{~km}=824.7 \times 10^{9} \mathrm{~m}$
Angular diameter $=35.72^{\prime \prime}=35.72 \times 4.874 \times 10^{-6} \mathrm{rad}$
Diameter of Jupiter = d
Using the relation,
$\theta=\frac{d}{D}$
$d=\theta D=824.7 \times 10^{9} \times 35.72 \times 4.872 \times 10^{-6}$
$=143520.76 \times 10^{3}=1.435 \times 10^{5} \mathrm{~km}$