Question

Plzz solve fast .because it's very urgent. Notethat in the question it is given diameter shrink not radius

Answer 1

Answer:

1%

Explanation:

we can assume this shrink as an error in measuring the  radius

note : if the diameter shrink = 1% , then the radius shrink = 0.5 %

g = GM/R²

log g = log G + log M - 2 logR

Δg/g = - 2ΔR/R

(Δg/g) x100%= (- 2ΔR/R)x100% = - 2 x 0.5% = 1%

% change = 1%