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

50 arcsec.

Step-by-step explanation:

Diameter of the Jupiter is 139822 km.

Minimum distance of the Jupiter from the earth is,

`S=588* 10^(6)km`

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

`\theta=(D)/(S)`

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

Now,

`\theta=(139822 km)/(588* 10^(6)km)\\\theta=237.792517* 10^(-6) arcradian`

We know that,

`1radian= (180* 60* 60)/(\pi ) arcsec`

Therefore,

`\theta=237.792517* 10^(-6)((180* 60* 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.