Question

Eris, the largest dwarf planet known in the Solar System, has a radius \(R=1200 \mathrm{~km}\) and an acceleration due to gravity on its surface of magnitude \(g=0.77 \mathrm{~m} / \mathrm{s}^{2}\) a) Use these numbers to calculate the escape speed from the surface of Eris. b) If an object is fired directly upward from the surface of Eris with half of this escape speed, to what maximum height above the surface will the object rise? (Assume that Eris has no atmosphere and negligible rotation.

Short answer

Question: Calculate the escape speed from the surface of Eris and find the maximum height above the surface an object will rise if it is fired directly upward from Eris with half of the escape speed. Answer: The escape speed from the surface of Eris is approximately 1200 m/s. An object fired directly upward from Eris with half of the escape speed will reach a maximum height of about 282,000 m above the surface.

Step-by-step solution

Write down the escape speed formula.

The escape speed formula is given by \(v_{e} = \sqrt{\frac{2GM}{R}}\), where \(v_{e}\) is the escape speed, \(G\) is the gravitational constant, \(M\) is the mass of the celestial body, and \(R\) is its radius.

Express the mass M in terms of acceleration due to gravity (g) and radius (R).

We can use the formula \(g = \frac{GM}{R^2}\), and then solving for M, we get \(M = \frac{gR^2}{G}\).

Substitute the expression for mass M into the escape speed formula.

Now we can rewrite the escape speed formula using the expression for mass M in terms of acceleration due to gravity (g) and radius R: \(v_{e} = \sqrt{\frac{2G\frac{gR^2}{G}}{R}}\).

Calculate the escape speed.

We are given the values of radius R (\(1200 \mathrm{~km}\)) and acceleration due to gravity g (\(0.77 \mathrm{~m} / \mathrm{s}^{2}\)). Don't forget to convert the radius from km to m. Now, we can plug these values into the escape speed formula: \(v_{e} = \sqrt{\frac{2G\frac{(0.77 \mathrm{~m} / \mathrm{s}^{2})(1200000 \mathrm{~m})^2}{G}}{1200000 \mathrm{~m}}}\) = \(\sqrt{2(0.77 \mathrm{~m} / \mathrm{s}^{2})(1200000 \mathrm{~m})}\) = \(1200 \mathrm{~m/s}\). So the escape speed from the surface of Eris is \(1200 \mathrm{~m/s}\). #b) Find the maximum height above the surface the object will rise.#

Write down the conservation of mechanical energy formula.

The conservation of mechanical energy formula is given by \(E_{1} = E_{2}\), where \(E_{1}\) is the initial total mechanical energy, and \(E_{2}\) is the final total mechanical energy.

Calculate the initial total mechanical energy.

The object is fired directly upward from the surface of Eris with half of the escape speed. The initial kinetic energy is \(K_{1} = \frac{1}{2}m(v_{0})^2\), where \(v_{0} = \frac{1}{2} \cdot 1200 \mathrm{~m/s} = 600 \mathrm{~m/s}\), and initial gravitational potential energy is \(U_{1} = -\frac{GMm}{R}\). The total initial mechanical energy is \(E_{1} = K_{1} + U_{1}\).

Calculate the final total mechanical energy.

The final mechanical energy is when the object reaches the maximum height above the surface of Eris. At this point, the object's final kinetic energy is \(K_{2} = 0\), as the object is momentarily at rest. The final gravitational potential energy is \(U_{2} = -\frac{GMm}{R+h}\), where \(h\) is the maximum height above the surface. The total final mechanical energy is \(E_{2} = K_{2} + U_{2}\).

Find the maximum height above the surface.

Using the conservation of mechanical energy formula, we can write: \(E_{1} = E_{2} \Rightarrow K_{1} + U_{1} = K_{2} + U_{2} \Rightarrow \frac{1}{2}m(600 \mathrm{~m/s})^2 - \frac{GMm}{1200000 \mathrm{~m}} = 0 - \frac{GMm}{R+h}\). Solving for \(h\), we have: \(h = \frac{GMm}{-\frac{1}{2}m(600 \mathrm{~m/s})^2 + \frac{GMm}{1200000 \mathrm{~m}}} - R = \frac{GMm}{-\frac{1}{2}(600 \mathrm{~m/s})^2 + \frac{GMm}{1200000 \mathrm{~m}}} - 1200000 \mathrm{~m}\). Plugging in the given values, we get: \(h \approx 282000 \mathrm{~m}\). So, the object will rise to a maximum height of about \(282000 \mathrm{~m}\) above the surface of Eris.

Key concepts

Escape Velocity

Escape velocity is the minimum speed needed for an object to break free from the gravitational pull of a celestial body without any further propulsion. This means that an object must be launched at or above this speed to completely escape the gravitational field. For planets and celestial bodies, escape velocity is an important concept because it determines whether an object will stay in orbit or fly off into space.
For Eris, the escape velocity can be calculated using the formula:

• \[ v_{e} = \sqrt{\frac{2GM}{R}} \]

This formula shows that escape velocity depends on the gravitational constant \(G\), the mass \(M\) of the celestial body, and its radius \(R\). As demonstrated in the solution, for Eris, escape velocity is about \(1200 \, \text{m/s}\).
Ultimately, this value tells us how fast an object needs to be moving if it wants to fully leave Eris's gravitational influence.

Conservation of Energy

The principle of conservation of energy states that the total energy in a closed system remains constant over time. In simpler terms, energy can change forms but is never lost or gained. This is fundamental when calculating how far or high something will travel in a system where gravity is involved. The total mechanical energy of an object is the sum of its kinetic energy (energy due to motion) and gravitational potential energy (energy due to position).
In the context of Eris, if an object is fired upward with half the escape velocity, we can use conservation of energy to find how high it will rise:

• Initial total energy (\( E_1 \)) = Kinetic energy + Initial gravitational potential energy
• Final total energy (\( E_2 \)) = Final gravitational potential energy (since kinetic energy is zero at the peak)

By setting these equal (\( E_1 = E_2 \)), we can solve for the maximum height. This ensures that all the initial kinetic energy becomes gravitational potential energy when the object reaches its peak height.

Gravitational Potential Energy

Gravitational potential energy (GPE) is the energy held by an object because of its position relative to a gravitational source, such as a planet or a moon. The higher an object is located in a gravitational field, the more gravitational potential energy it has. In terms of equations, GPE is given by:

• \[ U = -\frac{GMm}{r} \]

Here, \(U\) represents gravitational potential energy, \(G\) is the universal gravitational constant, \(M\) is the mass of the celestial body, \(m\) is the mass of the object, and \(r\) is the distance from the center of mass (for Eris, \(r = R + h\)).
This formula accounts for how higher altitudes lead to higher GPE, while the negative sign indicates that gravitational forces are attractive. Putting it into practice, when an object is projected with a certain kinetic energy, this energy is transformed into potential energy as it rises until it reaches maximum height, where all kinetic energy is converted into potential energy.