Question

Jupiter's moon Io has active volcanoes (in fact, it is the most volcanically active body in the solar system) that eject material 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.
Part A
How high would this material go on earth if it were ejected with the same speed as on Io? (RE = 6370 km, mE=5.96×1024kg)
Express your answer with the appropriate units.
View Available Hint(s)for Part A
Hint 1for Part A. Find speed of the material ejected from Io.
Step 1: Use conservation of energy (universal potential energy) to find the speed with which the material is ejected from Io. Remember that at the maximum altiute reached bu the material the speed is zero.

Step 2: After you find the speed of ejection from step 1, use the conservation of energy for the material ejected with this speed from Earth and find the altitude (distance from surface).

Answer 1

The conservation of energy allows us to find the speed of the lava expelled from Io and if at that speed, what height is expelled on Earth are:

    a) The velocity of lava expulsion at Io is: v = 1,168 m / s

    b) With this initial velocity the lava is not expelled from the Earth.

The conservation of energy is one of the fundamental principles of physics, stable that the mechanical energy formed by the kinetic energy and the potential energies is constant at all points if there is no friction in the system.

          Em = K + U

The kinetic energy is

          K = ½ m v²

The gravitational potential energy is:

          U =$-G \frac{Mm}{r^2}$

Where G is the universal gravitational constant, m is the mass of the lava, M the mass of the satellite, r the distance from the center of the satellite.

 

In this case we are looking for energy at two points.

Starting point. When the material is ejected.

          $Em_o$ = K + U

          $Em_o$ = ½ mv² - $G \frac{Mm}{R^2}$  

Final point. The highest point of the trajectory.

           $Em_f$ = U

           $Em_f = - G \frac{Mm}{( R+h)^2}$

Where h is the height to which the lava rises

          h = 500 km = 500 10³ m = 0.5 10⁶ m

Since there is no friction, the mechanical strength is preserved.

           $Em_o = Em_f$  

           $\frac{m v^2}{2} - G \frac{Mm}{R^2} = - G \frac{Mm}{(R+h)^2}$  

           

          $v^2 = 2GM ( \frac{1}{R^2} - \frac{1}{(R+h)^2} )$  

Let's calculate

         v² = 2 6.67 10⁻¹¹ 8.93 10²² ( ${ \frac{1}{1.821^2 } - \frac{1}{(1.821 +0.5)^2} } )10^{-12}$

         v² = 1.19126 10¹³ (0.11456 10⁻¹²) = 1.3646

         v = 1,168 m / s

Now let's find the height of the matter if it was expelled on Earth.

Starting point. At the surface.

        Em₀ = K + U

Final point. At the highest point

        $Em_f$ = U

Energy is conserved

        Em₀ = $Em_f$  

        $\frac{mv^2}{2} - G \frac{Mm}{R^2} = -G \frac{Mm}{(R+y)^2 }$  

         $\frac{1}{(R+y)^2 } = (G \frac{M}{R^2} - \frac{v^2}{2} ) \ \frac{1}{GM}$

         

Let's calculate

         $\frac{1}{(R+y)^2} = ( \frac{6.67 \ 10^{-11} \ 5.96 \ 10^{24} } {(6.370 \ 10^6)^2} - \frac{1.168^2}{2} ) \frac{1}{6.67 \ 10^{-11} \ 5.96 \ 10^{24} }$

         $\frac{1}{(R+y)^2} = \frac{9.797-0.6821}{39.75}\\ (R+y) = \sqrt{4.361}\\ R+y = 2.09 m$

We can see that with this exit velocity, the lava cannot be expelled from the planet's surface.

In conclusion, using the conservation of energy we can find the speed of the lava expelled from Io and if at that speed, what height is lava expelled on Earth are:

    a) The velocity of lava expulsion at Io is: v = 1,168 m / s

    b) With this initial velocity the lava is not expelled from the Earth.

Learn more here:  brainly.com/question/2737162