Question

What is the angular diameter of planet Jupiter as seen from Earth? We know that the diameter of Jupiter is and the minimum distance from Earth to Jupiter is . 25 arcsec 50 arcsec 60 arcsec 0.5 arcsec

Answer 1

Answer:

50 arcsec.

Explanation:

Diameter of the Jupiter is 139822 km.

Minimum distance of the Jupiter from the earth is,

$S=588\times 10^{6}km$

Now angular diameter of the Jupiter from the Earth can be calculate as,

$\theta=\frac{D}{S}$

Here, D is the diameter, and S is the minimum distance from the observing planet.

Now,

$\theta=\frac{139822 km}{588\times 10^{6}km}\\\theta=237.792517\times 10^{-6} arcradian$

We know that,

$1radian= \frac{180\times 60\times 60}{\pi } arcsec$

Therefore,

$\theta=237.792517\times 10^{-6}(\frac{180\times 60\times 60}{\pi } arcsec)\\\theta=49.07arcsec$

Therefore, the angular diameter of planet Jupiter is 49.07 arcsec which is quite close to the 50 arcsec which is given in options.