Question

A spherical asteroid has a radius of \(365.1 \mathrm{~km}\). The escape speed from its surface is \(319.2 \mathrm{~m} / \mathrm{s}\). What is the mass of the asteroid?

Short answer

Answer: Approximately \(7.88 \times 10^{20} \mathrm{kg}\).

Step-by-step solution

Write down the escape velocity formula

The escape velocity formula is given by: \(v_e = \sqrt{\frac{2GM}{r}}\) Where: - \(v_e\) is the escape velocity (319.2 m/s) - \(G\) is the gravitational constant (\(6.674 \times 10^{-11} \mathrm{m}^3 \mathrm{kg}^{-1} \mathrm{s}^{-2}\)) - \(M\) is the mass of the asteroid (our unknown) - \(r\) is the radius of the asteroid (365.1 km)

Convert radius to meters

Since the given radius is in kilometers and the other units are in meters, we need to convert the radius into meters: \(r_{m} = 365.1 \mathrm{~km} \times 1000\mathrm{~m/km} = 365100 \mathrm{~m}\)

Rearrange the formula to find the mass

First, we need to isolate the mass (M) in the escape velocity formula. By squaring both sides of the equation and multiplying by r, we get: \(M = \frac{v_e^2 r}{2G}\)

Substitute the given values and calculate the mass

Now we can plug in the given values and make the calculation: \(M = \frac{(319.2 \mathrm{~m/s})^2 × 365100 \mathrm{~m}}{2 \times 6.674 \times 10^{-11} \mathrm{m}^3 \mathrm{kg}^{-1} \mathrm{s}^{-2}}\) \(M \approx 7.88 \times 10^{20} \mathrm{kg}\) So, the mass of the asteroid is approximately \(7.88 \times 10^{20} \mathrm{kg}\).

Key concepts

Gravitational Constant

Understanding the gravitational constant (\textbf{G}) is crucial for grasping many concepts in physics, especially when dealing with gravitational forces between two bodies. This fundamental constant represents the force of attraction between two masses that are one meter apart. It's value is extremely small, approximately \(6.674 \times 10^{-11} \mathrm{m}^3 \mathrm{kg}^{-1} \mathrm{s}^{-2}\), indicating that gravity is a relatively weak force.
The gravitational constant is a key part of Newton's law of universal gravitation, which forms the foundation of celestial mechanics and astrophysics. This constant allows us to calculate the force of attraction between celestial bodies like planets, stars, and in our problem, asteroids. Given its universality, \(G\) is not just a figure to memorize but an essential component to understand the mechanics of our universe.
When calculating astronomical masses or escape velocities, \(G\) allows us to work with the enormous scales of space, despite its relatively minuscule numerical value. That's the beauty of universal constants: they provide a stable framework that applies regardless of where you are in the universe.

Asteroid Mass Calculation

Calculating the mass of an asteroid, like any celestial object, involves applying the laws of gravity and motion. Asteroids don't have paths around a star that are as clearly defined as planets, which makes direct observation methods more challenging.
Instead, we often rely on the concept of escape velocity—the minimum speed needed for an object to break free from another object's gravitational pull without further propulsion. This can be measured or estimated based on observations of objects in the vicinity of the asteroid or from the movement of a spacecraft that passes by or lands on it.
By using the escape velocity, we can infer the asteroid's mass. Given the escape velocity formula \(v_e = \sqrt{\frac{2GM}{r}}\), we rear back to our fundamental constant \(G\) and the measured radius \(r\) of the asteroid. Once escape velocity \(v_e\) is known, we can rearrange the formula to solve for mass \(M\), illuminating the otherwise elusive characteristic of the asteroid's mass.

Escape Velocity Formula

The escape velocity formula is an astonishing expression that tells us the speed needed to break the gravitational bind of a celestial body without further propulsion. Its formula, \(v_e = \sqrt{\frac{2GM}{r}}\), elegantly encapsulates the interplay between mass (\textbf{M}), radius (\textbf{r}), and the gravitational constant (\textbf{G}).
To understand this formula, it's helpful to recognize that escape velocity depends on two main factors: the mass of the body you're escaping from, and how far away from it you are. A larger mass or a closer proximity requires a higher escape velocity.
The beauty of this formula is that it does not depend on the mass or shape of the object that's trying to escape, but solely on the characteristics of the body being escaped from. When students learn to use this formula by substituting the known values, such as in our exercise where we calculated the asteroid's mass, they take an important step in mastering gravitational concepts important for understanding orbits, space travel, and the dynamics of our cosmos.