The more powerful the gravitational force of a planet, the greater its escape speed, \(v,\) and the greater the gravitational acceleration, \(g\), at its surface. However, in Table 12.1 , the value for \(v\) is much greater for Uranus than for Earth - but \(g\) is smaller on Uranus than on Earth! How can this be?

Short Answer

Expert verified
Question: Explain why the escape speed for Uranus is greater than that for Earth, but the gravitational acceleration is smaller on Uranus than on Earth. Answer: The escape speed of Uranus is greater than that of Earth because the mass of Uranus is much greater, while its radius is also significantly larger, resulting in a higher value in the escape speed formula. However, the gravitational acceleration on Uranus is smaller than on Earth because the mass-to-radius ratio is lower for Uranus, leading to a lower value in the gravitational acceleration formula.

Step by step solution

01

Gravitational acceleration formula

The formula for gravitational acceleration (g) at the surface of a planet is given by: g = G * (M / R^2) where G is the gravitational constant (6.674 × 10^-11 N m²/kg²), M is the mass of the planet, and R is the radius of the planet.
02

Escape speed formula

The formula for escape speed (v) is given by: v = sqrt(2 * G * M / R) where G is the gravitational constant, M is the mass of the planet, and R is the radius of the planet.
03

Escape speed and gravitational acceleration relationship

Notice that both formulas have a ratio of mass to radius, which is important for understanding the given observation. If a planet has a high mass and a low radius, its gravitational acceleration is high. If a planet has a high mass but a large radius, its escape speed is relatively high.
04

Compare Earth and Uranus

We know from the exercise that the escape speed (v) for Uranus is greater than for Earth, and the gravitational acceleration (g) is smaller on Uranus than on Earth. To understand this observation, let's compare the mass and radius of Earth and Uranus. Mass of Earth (M_E) = 5.97 × 10^24 kg Radius of Earth (R_E) = 6,371,000 m Mass of Uranus (M_U) = 86.8 × 10^24 kg Radius of Uranus (R_U) = 25,362,000 m
05

Analyze mass to radius ratios

Let's analyze the mass-to-radius ratios for both planets: M_E / R_E^2 = (5.97 × 10^24 kg) / (6,371,000 m)^2 = 1473.1 kg/m^2 M_U / R_U^2 = (86.8 × 10^24 kg) / (25,362,000 m)^2 = 1075.7 kg/m^2 Since the mass-to-radius ratio is higher for Earth than for Uranus (1473.1 > 1075.7), we can conclude that the gravitational acceleration (g) is greater on Earth than on Uranus.
06

Analyze escape speed factors

Now let's analyze the escape speed factors: sqrt(M_E / R_E) = sqrt((5.97 × 10^24 kg) / (6,371,000 m)) = 1.08 × 10^13 kg/m^3 sqrt(M_U / R_U) = sqrt((86.8 × 10^24 kg) / (25,362,000 m)) = 1.62 × 10^13 kg/m^3 Since the square root of the mass-to-radius ratio is higher for Uranus than for Earth (1.62 × 10^13 > 1.08 × 10^13), we can conclude that the escape speed (v) is greater for Uranus than for Earth. In summary, the escape speed of Uranus is greater than that of Earth because the mass of Uranus is much greater, while its radius is also significantly larger - leading to a higher value in the escape speed formula. At the same time, the gravitational acceleration on Uranus is smaller than on Earth because the mass-to-radius ratio is lower for Uranus, which results in a lower value in the gravitational acceleration formula.

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Most popular questions from this chapter

A satellite of mass \(m\) is in an elliptical orbit (that satisfies Kepler's laws) about a body of mass \(M,\) with \(m\) negligible compared to \(M\) a) Find the total energy of the satellite as a function of its speed, \(v\), and distance, \(r\), from the body it is orbiting. b) At the maximum and minimum distance between the satellite and the body, and only there, the angular momentum is simply related to the speed and distance. Use this relationship and the result of part (a) to obtain a relationship between the extreme distances and the satellite's energy and angular momentum. c) Solve the result of part (b) for the maximum and minimum radii of the orbit in terms of the energy and angular momentum per unit mass of the satellite. d) Transform the results of part (c) into expressions for the semimajor axis, \(a\), and eccentricity of the orbit, \(e\), in terms of the energy and angular momentum per unit mass of the satellite.

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