# Gravitational Constant and Velocity

• DameLight
In summary, Jupiter's moon Io has active volcanoes that can eject material up to 300 km above its surface. Using the conservation of mechanical energy, the material on Earth would reach a height of approximately 6.37 x 10^6 m if ejected with the same speed as on Io. However, this is likely an overestimation since it does not take into account the variation in gravity. A more accurate value can be obtained by using the approximation that the force of gravity is constant, which gives a height of approximately 1.25 x 10^5 m. It can also be noted that the gravity on Io is less than on Earth, which can be verified using the expression Ug = GMm/r and calculating the
DameLight

## Homework Statement

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

Ug = - (GMm)/r
Fg = (GMm)/r2

## 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:
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?

DameLight
DameLight said:
- (GMm)/r = 1/2 mv2
- (GM)/r = 1/2 v2
- 2(GM)/r = v2
v = √(- 2(GM)/r)
r = rIo + h
v = √(- 2(6.67 × 10−11*8.94 × 1022)/(1815 * 103 + 300 * 103))
v = 2380 m/s
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).

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)

Thank you for pointing that out : )

DameLight said:
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)
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.

DameLight
DameLight said:

## Homework Statement

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

Ug = - (GMm)/r
Fg = (GMm)/r2

## 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

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?

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.

## 1. What is the gravitational constant?

The gravitational constant, denoted as G, is a fundamental physical constant that represents the strength of the gravitational force between two objects with mass. It is a universal constant and is a key component in the formula for calculating the force of gravity between two bodies.

## 2. How is the gravitational constant measured?

The value of the gravitational constant is determined through experiments and observations of the gravitational force between objects with known masses and distances. Various methods, such as the Cavendish experiment, are used to measure G with high precision.

## 3. What is the relationship between the gravitational constant and velocity?

The gravitational constant does not directly depend on velocity. However, velocity does play a role in the formula for calculating the force of gravity, as it is used to determine the acceleration due to gravity, which is then multiplied by the masses of the objects and the gravitational constant.

## 4. How does the gravitational constant affect the motion of celestial bodies?

The gravitational constant plays a crucial role in determining the trajectory and dynamics of celestial bodies. It is responsible for the orbits of planets around the sun, the motion of satellites around Earth, and the interactions between galaxies in the universe.

## 5. Has the value of the gravitational constant changed over time?

Currently, the value of the gravitational constant is considered to be a constant in the standard model of physics. However, there have been some discrepancies in experimental measurements, leading to the possibility that the value of G may change over time. Research and experiments are ongoing to further understand the nature of this fundamental constant.

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