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Ksenya-84 [330]
3 years ago
13

Jupiter's moon Io has active volcanoes (in fact, it is the most volcanically active body in the solar system) that eject materia

l as high as 500 km (or even higher) above the surface. Io has a mass of 8.93×10²²kg and a radius of 1821 km. For this calculation, ignore any variation in gravity over the 500 km range of the debris. How high would this material go on earth if it were ejected with the same speed as on Io?
Physics
1 answer:
melisa1 [442]3 years ago
5 0

Answer:

H_2 = 91.55 km

Explanation:

Gravity on the surface of planet is given as

g = \frac{GM}{R^2}

as we know that

M = 8.93 \times 10^{22} kg

R = 1821 km

now gravity on the planet is

g = \frac{(6.67 \times 10^{-11})(8.93 \times 10^{22})}{(1821 \times 10^3)^}

so we have

g = 1.8 m/s^2

now we know that

H_{max} = \frac{v^2}{2g}

so we will say

\frac{H_1}{H_2} = \frac{g_2}{g_1}

\frac{500}{H_2} = \frac{9.81}{1.8}

H_2 = 91.55 km

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Two planets P1 and P2 orbit around a star S in circular orbits with speeds v1 = 40.2 km/s, and v2 = 56.0 km/s respectively. If t
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Answer: 3.66(10)^{33}kg

Explanation:

We are told both planets describe a circular orbit around the star S. So, let's approach this problem begining with the angular velocity \omega of the planet P1 with a period T=750years=2.36(10)^{10}s:

\omega=\frac{2\pi}{T}=\frac{V_{1}}{R} (1)

Where:

V_{1}=40.2km/s=40200m/s is the velocity of planet P1

R is the radius of the orbit of planet P1

Finding R:

R=\frac{V_{1}}{2\pi}T (2)

R=\frac{40200m/s}{2\pi}2.36(10)^{10}s (3)

R=1.5132(10)^{14}m (4)

On the other hand, we know the gravitational force F between the star S with mass M and the planet P1 with mass m is:

F=G\frac{Mm}{R^{2}} (5)

Where G is the Gravitational Constant and its value is 6.674(10)^{-11}\frac{m^{3}}{kgs^{2}}

In addition, the centripetal force F_{c} exerted on the planet is:

F_{c}=\frac{m{V_{1}}^{2}}{R^{2}} (6)

Assuming this system is in equilibrium:

F=F_{c} (7)

Substituting (5) and (6) in (7):

G\frac{Mm}{R^{2}}=\frac{m{V_{1}}^{2}}{R^{2}} (8)

Finding M:

M=\frac{V^{2}R}{G} (9)

M=\frac{(40200m/s)^{2}(1.5132(10)^{14}m)}{6.674(10)^{-11}\frac{m^{3}}{kgs^{2}}} (10)

Finally:

M=3.66(10)^{33}kg (11) This is the mass of the star S

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