Question

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

Acceleration of free fall towards the sun at the distance of earth's orbit is $5.9\times {10}^{-3}\mathrm{m}/{\mathrm{s}}^2$.

Step-by-step solution

Given information 

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

Explanation

WE know, $F=G\frac{M\text{sun}m}{R^2}$

Again, $F=m{g}_{\text{sun}}$

So

$m{g}_{sun}=G\frac{{{\displaystyle M}}_{sun}m}{R^2}$

$g_{sun}=\frac{M_{sun}G}{R^2}$

Acceleration of free fall

Here,

$M_{\text{sun}}=1.99\times {20}^{30}\mathrm{kg}$

localid="1649396961498" $G=6.67\times {10}^{-11}{\mathrm{Nm}}^2/k{g}^2$

And $R=R_{\text{sun-earth}}-R_{\text{earth}}$

localid="1649396020677" $=\left(1.5\times {10}^{11}m-6.4\times {10}^{6}m\right)$

localid="1649395379454" $=1.499936\times {10}^{11}\mathrm{m}$

localid="1649395636670" $g_{\text{sun}}=\left(\frac{1.99\times {10}^{30}kg\times 6.67\times {10}^{-11}N{m}^2/k{g}^2}{{\left(1.499936\times {10}^{11}m\right)}^2}\right)$

$=5.9\times {10}^{-3}\mathrm{m}/{\mathrm{s}}^2$

Final answer

Acceleration of free fall towards the sun at the distance of earth's orbit is $5.9\times {10}^{-3}\mathrm{m}/{\mathrm{s}}^2$.