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

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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

01

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.
02

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 \(M = \frac{gR^2}{G}\).
03

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: \(v_{e} = \sqrt{\frac{2G\frac{gR^2}{G}}{R}}\).
04

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.#
05

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.
06

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 \(E_{1} = K_{1} + U_{1}\).
07

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 \(E_{2} = K_{2} + U_{2}\).
08

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.

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