Does the acceleration due to gravity (9.80m/s2) dependent upon the mass of an object.?

No. it's a fixed value for planet earth !

Galileo's experiments already proved that mass is independent of rate of fall.

ma=mg

a=g

Weight as in W = mg is mass m circumstances a g value that has the contraptions m/sec^2, that are the contraptions for acceleration. So we call g the acceleration by using gravity. yet, and it somewhat is a super yet, that is not consistent as you recommend. g = GM/R^2 actually; the place G is a real consistent, M is the mass source of the gravity field, and R is the midsection to midsection distance between despite mass is being weighed m and the source mass M. If r is Earth's radius and h is the peak of m above floor, then R = r + h. As you will see if h gets extra advantageous and the gap above floor gets bigger, R additionally gets extra advantageous. this suggests that g gets smaller considering M is Earth's mass and caught. And there you're, g varies "in line with its distance from the floor." Your confusion comes from poorly written textbooks who declare gravity acceleration g is persevering with. that is not. g is persevering with in simple terms for a fastened place R and mass M...otherwise g varies. while i take advantage of g = 9.80 one m/sec^2 in my solutions, that's a user-friendly form we see in textbooks, I almost consistently specify "close to Earth's floor," which fixes R and M for a given question. wherein case, g is extremely 9.80 one m/sec^2, yet in simple terms for those circumstances. in case you have been to crunch g = GM/R^2 for the Moon mass M and radius R, you will locate that g is a few million/6 that of Earth's g. So in case you're slightly obese on earth, you're able to weigh in at a million/6 your Earth weight on the Moon.