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

a) Calculate the escape speed: First, we find the mass (M) of Eris using the acceleration due to gravity formula: \(M = \frac{0.77 \mathrm{~m} / \mathrm{s}^{2} \times (1200 \times 10^3 \mathrm{~m})^2}{6.674 \times 10^{-11} \mathrm{m}^{3} \cdot \mathrm{kg}^{-1} \cdot \mathrm{s}^{-2}} = 1.43 \times 10^{22} \mathrm{kg}\) Now, we can use the escape speed formula: \(v_{e} = \sqrt{\frac{2(6.674 \times 10^{-11} \mathrm{m}^{3} \cdot \mathrm{kg}^{-1} \cdot \mathrm{s}^{-2})(1.43 \times 10^{22} \mathrm{kg})}{1200 \times 10^3 \mathrm{~m}}} = 841.59 \mathrm{~m} / \mathrm{s}\) b) Calculate the maximum height: Now let's find the maximum height (h) an object will rise when fired with half the escape speed. Using the conservation of mechanical energy formula, and canceling out mass (m), we rearrange the equation to find height (h): \(h=\frac{R(1 - \frac{1}{4})}{\frac{1}{4}} - R = \frac{3}{4}R\) Calculate the value of h using the radius obtained in part a): \(h = \frac{3}{4} (1200 \times 10^3 \mathrm{~m}) = 900 \times 10^3 \mathrm{~m}\) So, the escape speed from the surface of Eris is approximately \(841.59 \mathrm{~m} / \mathrm{s}\), and the maximum height an object will rise when fired directly upward with half the escape speed is \(900 \times 10^3 \mathrm{~m}\).

Step-by-step solution

a) Calculate the escape speed

First, let's find the mass (M) of Eris using the acceleration due to gravity formula: \(g = \frac{GM}{R^2}\) Now, we re-arrange the formula to find mass (M): \(M = \frac{gR^2}{G}\) Now we can plug in the values to find the mass: \(M = \frac{0.77 \mathrm{~m} / \mathrm{s}^{2} \times (1200 \times 10^3 \mathrm{~m})^2}{6.674 \times 10^{-11} \mathrm{m}^{3} \cdot \mathrm{kg}^{-1} \cdot \mathrm{s}^{-2}}\) Calculate the escape speed formula: \(v_{e} = \sqrt{\frac{2GM}{R}}\) By putting values in the escape speed formula: \(v_{e} = \sqrt{\frac{2(6.674 \times 10^{-11} \mathrm{m}^{3} \cdot \mathrm{kg}^{-1} \cdot \mathrm{s}^{-2})(M)}{1200 \times 10^3 \mathrm{~m}}}\) Calculating the value of \(v_{e}\).

b) Calculate the maximum height

Now, let's find the maximum height (h) an object will rise when fired with half the escape speed. We will use the conservation of mechanical energy formula: \(E_{initial} = E_{final}\) The initial energy consists of kinetic and potential energy: \(E_{initial} = \frac{1}{2}mv^2 - \frac{GMm}{R}\) The final energy is the potential energy when it reaches the highest point (h): \(E_{final} = -\frac{GMm}{R+h}\) Now set initial energy equal to the final energy and solve for height (h): \(\frac{1}{2}mv^2 - \frac{GMm}{R} = -\frac{GMm}{R+h}\) The object is fired with half the escape speed, so \(v=v_{e}/2\). Cancel out mass (m) and rearrange the equation to find height (h): \(h=\frac{R(1 - \frac{1}{4})}{\frac{1}{4}} - R\) Calculate the value of h using the escape speed and radius obtained in part a).

Key concepts

Gravitational Physics

Gravitational physics is a fundamental branch of physics that deals with the force of attraction between two masses. It's based on Newton's universal law of gravitation, which states that every point mass attracts every other point mass with a force that is proportional to the product of their masses and inversely proportional to the square of the distance between their centers.

This concept is crucial when calculating the escape speed from a celestial body, like the dwarf planet Eris. The escape speed is the minimum velocity an object needs to be given to 'break free' from the gravitational pull of the planet or moon, without further propulsion. Mathematics plays a key role here, as the escape speed () is derived from the gravitational constant (), the mass of the celestial body (), and its radius ().Using the formula , we can determine the initial force and speed needed to overcome gravity's bond. The formula itself is a practical demonstration of gravitational physics, linking the forces and distances involved in celestial mechanics to tangible numbers that can be calculated and understood.

Conservation of Mechanical Energy

The principle of conservation of mechanical energy states that the total mechanical energy in a system remains constant if only conservative forces, like gravity, are doing work. Mechanical energy itself is the sum of kinetic energy (due to movement) and potential energy (due to position).When applied to an escaping object from a planet like Eris, we look at how its kinetic energy and gravitational potential energy change as it moves away from the planet. Initially, the object has a certain amount of kinetic energy, given by , and gravitational potential energy, given by . As the object ascends, it loses kinetic energy and gains potential energy.When analyzing an object projected upward from Eris with half the escape speed, we can predict its maximum height by setting the initial mechanical energy equal to the mechanical energy at the highest point. Through this balance, we can solve for the height () the object reaches before it comes to a stop and begins to fall back down to Eris. This is a beautiful illustration of energy conservation in action and drives home the predictability and consistency of physical laws in gravitational fields.

Dwarf Planet Mass Calculation

Calculating the mass of a dwarf planet like Eris involves understanding the relationship between gravity and mass. Dwarf planets, despite being smaller than the 'official' planets, still exert sufficient gravitational force to shape themselves into a nearly round form. To calculate their mass, we rely on the formula derived from the second law of Newton applied to gravitational force: .From this relationship, we re-arrange the terms to solve for the mass () when the acceleration due to gravity at the surface () and the radius () are known. With being the universal gravitational constant, one can insert the known values and perform the calculation to find the mass.This process is not only fundamental in understanding the size and composition of celestial bodies but also critical for planning space missions and predicting orbits. Working through such a mass calculation gives students a practical insight into the direct connection between observable gravity on a cosmic scale and the underlying mass driving it.