Question

HEY MATES....

$when \: planet \: jupiter \: is \: at \: \\ distance \: of \: 824.7million \: km \\ of \: the \: earth \: it \: angular \: \\ diameter \: is \: measured \: to \: be \: \\ {35.27}^{ || } of \: arc \: calculate \: the \: \\ diameter \: of \: jupiter.....$
THANKS FOR UR EFFORTS ❤️☺️❤️​

Answer 1

Distance of Jupiter from the Earth

D = 824.7×10^6 km = 824.7 × 10^9 m

Angular diameter

θ = 35.72“ = 35.72×4.874×10^-6 rad

# Formula-

Angular diameter of jupiter is given by relation,

θ = d/D

d = θ.D

d = 824.7×10^9×35.72×4.872×10^-6

d = 143520.76×10^3 m

d = 1.435 × 10^5 Km

Hence, diameter of jupiter is 1.4×10^5 km.

hope it helps you mark as brainliest please

Answer 2

PRE - KNOWLEDGE REQUIRED

● to convert an angle given in arcsecond unit (") into radian unit (rad) we have to multiply the angle given in arcseconds by 4.874 × 10^(-6) .

● we should know the relation that,

theta = d / D

where theta is the angular diameter ; d is the diameter of planet ( body) and D is the distance between the planet X in the question and earth.

Now refer to the image attached