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LuckyWell [14K]
4 years ago
7

A person’s will be the same on Jupiter and Earth. A person’s will be greater on Jupiter than on Earth.

Biology
2 answers:
inysia [295]4 years ago
8 0

Answer:

A person will be greater on Jupiter than Earth. The surface gravity is having is larger than Earth’s by roughly 2.4 times. So, if you weigh 100 pounds on Earth it would be 240 on Jupiter. Hopefully this helps! Feel free to mark this as brainliest!  

koban [17]4 years ago
6 0

Answer:

Greater on Jupiter

Explanation:

Gravity depends on mass. Jupiter has more mass than Earth so its gravity is two and a half times greater. Therefore a person will weigh more on Jupiter.

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Mars had an orbital period of 1.88 years. In two or more complete sentences,explain how to calculate the average distance from m
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This is an interesting problem involving astronomy, in fact, simple physics.
Let r=distance of sun to mars, in metres

<span>Mars had an orbital period of 1.88 years.
=>
tangential velocity, v, of the planet, in m/s is
v=\frac{2\pi{r}}{T}
</span>=\frac{2\pi{r}}{1.88*365.25*86400} m/s, accounting for leap years
=3.371*10^{-8}\pi{r}   m/s

The centripetal force, Fc, generated is
Fc=\frac{mv^2}{r}   
        where m=mass of mars = 6.39*10^(24) kg
=\frac{mv^2}{r}
=\frac{6.39*10^{24}v^2}{r}
=7.26168*10^9\pi^2r

The gravitation pull from the sun, Fg, is given by
Fg=\frac{GMm}{r^2}
    where G=grav. const., =6.67408*10^(-11) m^3<span> kg^(</span>-1)<span> s^(</span>-2)
              M=mass of sun=1.989*10^(30) kg
=\frac{6.67408*10^{-11}1.989*10^{30}6.39*10^{24}}{r^2}
=\frac{8.4826*10^44}{r^2}

Since the radial distance is in equilibrium, the average distance, r can be found by equation Fc=Fg and solving for r:
Fc=Fg
=>
7.26168*10^9\pi^2r=\frac{8.4826*10^44}{r^2}
Solving for the real root:
r^3=\frac{8.48256*10^44}{7.26168*10^9*%pi^2}
=\frac{1.1681263*10^{35}}{\pi^2}
=2.279*10^11 m


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