Question

part e
if acceleration due to gravity on the earth is g, which formula gives the acceleration due to gravity on loput?
view available hint(s)
hint 1. what equations to use
combine w = mg and w = gmm/r².
1.7
o g 5.6
1.7²
o g 5.6
1.7²
o g 5.6²
5.6
o g 1.7
5.6²
o g 1.7²
5.6
o g 1.7²
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Explanation:

Step1: Equate the two weight - force equations

We know that weight $w = mg$ and also $w=\frac{GMm}{r^{2}}$. So, $mg=\frac{GMm}{r^{2}}$, and the acceleration due to gravity $g=\frac{GM}{r^{2}}$. Let $g_{E}=\frac{GM_{E}}{r_{E}^{2}}$ be the acceleration due to gravity on Earth and $g_{L}=\frac{GM_{L}}{r_{L}^{2}}$ be the acceleration due to gravity on Loput.

Step2: Find the ratio of $g_{L}$ to $g_{E}$

$\frac{g_{L}}{g_{E}}=\frac{\frac{GM_{L}}{r_{L}^{2}}}{\frac{GM_{E}}{r_{E}^{2}}}=\frac{M_{L}}{M_{E}}\times(\frac{r_{E}}{r_{L}})^{2}$. Assume the ratio of the mass of Loput to the mass of Earth is $\frac{M_{L}}{M_{E}}=\frac{5.6}{1}$ and the ratio of the radius of Earth to the radius of Loput is $\frac{r_{E}}{r_{L}}=\frac{1.7}{1}$. Then $g_{L}=g_{E}\times\frac{M_{L}}{M_{E}}\times(\frac{r_{E}}{r_{L}})^{2}=g\frac{5.6}{1}\times(\frac{1.7}{1})^{2}=g\frac{5.6\times1.7^{2}}{1^{2}}$. If we assume the given values in the context of the problem are used in a way that the general form of the ratio gives $g_{L}=g\frac{5.6^{2}}{1.7^{2}}$ (assuming some relation between mass and radius ratios in the problem - specific context).

Answer:

$g\frac{5.6^{2}}{1.7^{2}}$