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What is free-fall acceleration toward the sun at the distance of the earth’s orbit? Astronomical data are inside the back cover of the book .

Short Answer

Expert verified

Acceleration of free fall towards the sun at the distance of earth's orbit is 5.9×10-3m/s2.

Step by step solution

01

Given information 

Given in the question that, Astronomic data in the book,

02

Explanation

WE know, F=GMsunmR2

Again, F=mgsun

So

mgsun=GMsunmR2

gsun=MsunGR2

03

Acceleration of free fall

Here,

Msun=1.99×2030kg

localid="1649396961498" G=6.67×10−11Nm2/kg2

And R=Rsun-earth-Rearth

localid="1649396020677" =1.5×1011m-6.4×106m

localid="1649395379454" =1.499936×1011m

localid="1649395636670" gsun=1.99×1030kg×6.67×10−11Nm2/kg21.499936×1011m2

=5.9×10−3m/s2

04

Final answer

Acceleration of free fall towards the sun at the distance of earth's orbit is 5.9×10-3m/s2.

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