An astronomical object has its mass 4 times the mass of earth and radius half of the radius of
earth. If acceleration due to gravity at earth is g, find its value at the surface of the astronomical
object.
Answers
Answered by
48
let x be the astronomical object
mass of x (M) = 4 times of mass of earth
= 4 × 5.972 × 10 ^ 24 kg
= 23.888 × 10 ^ 24 kg
radius of x (r)= 1/2 of radius of earth
= 1/2 × 6371 km
= 3185.5 km
= 3185500 m
we know that acceleration due to gravity on x = GM/r^2
= (6.673 × 10^-11 Nm^2kg^-2× 23.888 × 10^24 kg)/(31855)m^2
= 39.27224 ms^-2
therefore acceleration due to gravity on the astronomical object = 39.27224 ms^-2
mass of x (M) = 4 times of mass of earth
= 4 × 5.972 × 10 ^ 24 kg
= 23.888 × 10 ^ 24 kg
radius of x (r)= 1/2 of radius of earth
= 1/2 × 6371 km
= 3185.5 km
= 3185500 m
we know that acceleration due to gravity on x = GM/r^2
= (6.673 × 10^-11 Nm^2kg^-2× 23.888 × 10^24 kg)/(31855)m^2
= 39.27224 ms^-2
therefore acceleration due to gravity on the astronomical object = 39.27224 ms^-2
Answered by
14
hope this helps
any confusion then ask me here or in inbox
hope this helps
any confusion then ask me here or in inbox
hope this helps
Attachments:
Mukundj:
among all answers yours is 100% correct, good job.
Similar questions