# Gravitational Constant and Velocity

1. Jun 26, 2015

### DameLight

1. The problem statement, all variables and given/known data
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 300 km (or even higher) above the surface. Io has a mass of 8.94 × 1022 kg and a radius of 1815 km . Ignore any variation in gravity over the 300 km range of the debris.

h = 300 km = 300 * 103 m
mIo = 8.94 × 1022 kg
rIo = 1815 km = 1815 * 103 m

How high would this material go on earth if it were ejected with the same speed as on Io?

ME = 5.97 × 1024
rE = 6371 km = 6371 * 103 m
G = 6.67 × 10−11

2. Relevant equations
Ug = - (GMm)/r
Fg = (GMm)/r2

3. The attempt at a solution
Conservation of ME on Io
PE = PE + KE

- (GMm)/(r+h) = - (GMm)/r + 1/2 mv2
- (GMm)/(r+h) + - (GMm)/r = 1/2 v2
2(- (GM)/(r+h) + - (GM)/r)) = v2

v = √(2(- (GM)/(r+h) + - (GM)/r)))
v = √(2(- (6.67 × 10−11 * 8.94 × 1022)/(1815 * 103 + 300 * 103) + - ((6.67 × 10−11 * 8.94 × 1022)/1815 * 103)))
v = 2563.36 m/s

Conservation of ME on Earth
PE = PE + KE
- (GMm)/(r+h) = - (GMm)/r + 1/2 mv2
(- (GMm)/r + 1/2 mv2)/- (GMm) = (r+h)
h = (- (GMm)/r + 1/2 mv2)/- (GMm) - r
h = (- (6.67 × 10−11 * 5.97 × 1024)/6371 * 103 + 1/2 * 2563.362)/- (6.67 × 10−11 * 5.97 × 1024) - 6371 * 103
h = 6.37 * 106

Points Changed:
1. Initial PE should exist

Last edited: Jun 27, 2015
2. Jun 26, 2015

### Simon Bridge

Initial kinetic energy changes the gravitational PE.
Change in PE is final minus initial ... where is the initial PE in each case?

Notice you are asked to ignore changes in gravity with height.
Did you do this?

3. Jun 26, 2015

### Nathanael

Your calculation of v is not right.

Your formula for gravitational potential, -GMm/r, takes the zero potential to be at infinity. What you really want is for the zero potential to be at the surface (that way your equation PE=KE will be true).

Anyway that expression for gravitational potential is the exact expression; it is the integral of the force, meaning it takes into account the variation of gravity. You can still use it if you want, but your answer will be more exact than the question expects (it tells you to ignore the variation of gravity).

4. Jun 26, 2015

### DameLight

Ah ok thank you : ) but I'm a little confused about the variation in gravity part can you explain it? (I just learned these yesterday)

5. Jun 26, 2015

### DameLight

Thank you for pointing that out : )

6. Jun 26, 2015

### Nathanael

The force of gravity varies with height, so it will not be the same the whole time the lava is in the air (or in the "no air").

The (change in) gravitational potential energy is the integral of the gravitational force with respect to distance. So if you want to find the exact (change in) potential energy, you will have to account for the way that gravity changes with height. That's where the -GMm/r comes from (if you know calculus: $GMm\int\frac{dx}{x^2}=-GMm(\frac{1}{x})+C$)

The problem simplifies this, though, by using an approximation that the force of gravity is constant. That way you can just find the force at the surface of the moon and then pretend like that is the force acting on the lava for the whole path. Then you can just use the formula: Δ[PE]=mgΔh (of course the "g" will not be the same as on Earth). This approximation will be pretty accurate, as long the lava doesn't go too high.

7. Jun 27, 2015

### PeroK

Do you believe your answer? The lava would go twenty times higher on Earth?

Do you think the gravity on Io is greater or less than on Earth? How could you check this?

8. Jun 27, 2015

### Simon Bridge

And still not following instructions: U=GMm/r is the wrong expression.

You know a simple approximation for Ug that uses g ... what is it?
You know what g is for the Earth, you can use that to find what g is for Io.