The mass of Venus is 81.5% that of the earth, and its radius is 94.9% that of th

smellmovinglz 2021-11-23 Answered
The mass of Venus is 81.5% that of the earth, and its radius is 94.9% that of the earth.
(a) Compute the acceleration due to gravity on the surface of Venus from these data.
(b) If a rock weighs 75.0 N on earth, what would it weigh at the surface of Venuse?

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Coldst
Answered 2021-11-24 Author has 7828 answers
(a) The mass of Venus \(\displaystyle{M}_{{{V}}}\) represent 81.5% of the of the Earth. So, the mass of Venus is
\(\displaystyle{M}_{{{V}}}\)=(81.5%)\(\displaystyle{M}_{{{E}}}\)
(1)
Where \(\displaystyle{M}_{{{E}}}\) is the mass of the Earth. Plug the mass of Earth into equation (1), so we can get \(\displaystyle{M}_{{{V}}}\) by
\(\displaystyle{M}_{{{V}}}={\left({0.815}\right)}{M}_{{{E}}}={\left({0.815}\right)}{\left({5.97}\times{10}^{{{24}}}{k}{g}\right)}={4.86}\times{10}^{{{24}}}{k}{g}\)
The radius of Venus represent 94.9% of the radius of the Earth. So, the radius of Venus is
\(\displaystyle{R}_{{{V}}}={\left({94.9}\%\right)}{R}_{{{E}}}\)
(2)
Where \(\displaystyle{R}_{{{E}}}\) is the radius of the Earth. Plug the radius of Earth into equation (2), so we can get \(\displaystyle{R}_{{{V}}}\) by
\(\displaystyle{R}_{{{V}}}={\left({0.949}\right)}{\left({6.38}\times{10}^{{{6}}}{m}\right)}={6.05}\times{10}^{{{6}}}{m}\)
The weight w of a body is the total gravitational force exerted on it by all other bodies in the univerce where this gravitational force F is given by Newton`s general law of gravity by
\(\displaystyle{F}={G}{\frac{{{M}_{{{V}}}{m}}}{{{{R}_{{V}}^{{2}}}}}}\)
(3)
Where \(\displaystyle{R}_{{{V}}}\) is the radius of Venus, m is the mass of the body. As we mentioned above, this gravitational force equals the weight of the body which is mg, so using equation (3) we get the acceleration on Venus by
\(\displaystyle{m}{g}={G}{\frac{{{M}_{{{V}}}{m}}}{{{{R}_{{V}}^{{2}}}}}}\)
\(\displaystyle{g}{v}={G}{\frac{{{M}_{{{V}}}}}{{{{R}_{{V}}^{{2}}}}}}\)
(4)
Now, we plug values for \(\displaystyle{R}_{{{V}}}\), \(\displaystyle{M}_{{{V}}}\) and G into equation (4) to get gv
\(\displaystyle{g}{v}={\frac{{{G}{M}_{{{V}}}}}{{{{R}_{{V}}^{{2}}}}}}\)
\(\displaystyle={\frac{{{\left({6.67}\times{10}^{{-{11}}}{N}\cdot{m}^{{{2}}}\div{k}{g}^{{{2}}}\right)}{\left({4.86}\times{10}^{{{24}}}{k}{g}\right)}}}{{{\left({6.05}\times{10}^{{{6}}}{m}\right)}^{{{2}}}}}}\)
\(\displaystyle={8.86}{m}\div{s}^{{{2}}}\)
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Supoilign1964
Answered 2021-11-25 Author has 6791 answers
(b) The mass doesn't change as the planet changes while the weight changes. So, we can get the mass of the rock on the Earth then use this value of mass to get the weight on Venus. On Earth, the weight is given by
\(\displaystyle{w}_{{{E}}}={m}_{{{g}{E}}}\)
\(\displaystyle{m}={\frac{{{w}_{{{E}}}}}{{{g}_{{{E}}}}}}\)
\(\displaystyle{m}={\frac{{{75}{N}}}{{{9.8}\frac{{m}}{{s}^{{{2}}}}}}}\)
\(\displaystyle{m}={7.65}{k}{g}\)
Now, we use the mass of the rock to get the weight of the rock on Venus using \(\displaystyle{g}_{{{V}}}\)
\(\displaystyle{W}_{{{V}}}={m}{g}_{{{V}}}={\left({7.65}{k}{g}\right)}{\left({8.86}\frac{{m}}{{s}^{{{2}}}}\right)}={67.8}{N}\)
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