Question

A spherical asteroid has a mass of \(1.869 \cdot 10^{20} \mathrm{~kg}\) and a radius of \(358.9 \mathrm{~km} .\) What is the escape speed from its surface?

Short answer

Answer: The escape speed from the surface of the asteroid is approximately \(263.53 \mathrm{m/s}\).

Step-by-step solution

Understanding the escape speed formula

Escape speed is the minimum speed required for an object to escape the gravitational influence of a massive body, without any additional propulsion. The escape speed formula can be written as: \(v_\text{escape} = \sqrt{\frac{2GM}{R}}\) where \(v_\text{escape}\) is the escape speed, \(G\) is the gravitational constant (\(6.67430 \times 10^{-11} \mathrm{m^3 kg^{-1} s^{-2}}\)), \(M\) is the mass of the asteroid, and \(R\) is the radius of the asteroid.

Convert the radius to meters

We are given the radius of the asteroid in kilometers, but we need to convert it to meters to have consistent units with the other values in the equation. To convert kilometers to meters, multiply by 1,000: \(R_\text{meters} = 358.9 \mathrm{km} \times 1,000 = 358{,}900 \mathrm{m}\)

Calculate the escape speed

Now we can plug in the values into the escape speed formula: \(v_\text{escape} = \sqrt{\frac{2 \times 6.67430 \times 10^{-11} \mathrm{m^3 kg^{-1} s^{-2}} \times 1.869 \times 10^{20} \mathrm{kg}}{358{,}900 \mathrm{m}}}\) Calculate the value inside the square root: \(v_\text{escape} = \sqrt{\frac{2.495 \times 10^{10} \mathrm{m^2 s^{-2}}}{358{,}900 \mathrm{m}}}\) Now, divide the numbers: \(v_\text{escape} = \sqrt{69{,}544.677 \mathrm{m^2 s^{-2}}}\) Finally, take the square root: \(v_\text{escape} = 263.53 \mathrm{m/s}\) So, the escape speed from the surface of the asteroid is approximately \(263.53 \mathrm{m/s}\).

Key concepts

Gravitational Constant

In the context of celestial mechanics and astrophysics, the gravitational constant (\( G \)) is a pivotal value in understanding the forces that govern the universe. It is defined as the proportionality factor used in the equation that relates the attractive force between two objects with mass. More specifically, it is the force per unit mass between two point masses separated by one meter. The value of the gravitational constant is approximately \(6.67430 \times 10^{-11} \text{m}^3 \text{kg}^{-1} \text{s}^{-2}\).

Understanding \( G \) is crucial not only in calculating escape speeds but also in determining forces, potential energy and dynamics of astronomical bodies’ movement. Gravitational constant allows us to appreciate the scale of gravitational influence not just on Earth, but across the cosmos, as it is a universal constant applicable everywhere in the universe.

Gravitational Influence

The concept of gravitational influence refers to the extent to which a celestial body, such as a planet, moon, or asteroid, can exert a force on objects within its vicinity due to its mass. This influence is why objects fall towards the center of the Earth when dropped, or why the moon orbits around the Earth. The strength of this influence decreases with the square of the distance between the masses involved, illustrating that gravity's reach is infinite but weakens significantly with distance.

When considering the gravitational influence of a body like an asteroid, it becomes apparent why calculating the escape speed is of interest. Escape speed signifies the velocity an object must reach to overcome the gravitational pull and travel away from the asteroid indefinitely without further propulsion - a fundamental concept for space travel and exploration.

Escape Velocity Calculation

Calculating the escape velocity is crucial for any space mission planning and understanding the dynamics of celestial bodies. Escape velocity refers to the minimum speed an object must achieve to break free from a celestial body’s gravity without additional thrust or fuel.

The calculation involves inputting values into the formula \(v_{\text{escape}} = \sqrt{\frac{2GM}{R}}\), where \(v_{\text{escape}}\) is the escape speed, \(G\) is the gravitational constant, \(M\) is the mass of the body being escaped from, and \(R\) is its radius. Typically, the radius should be converted into meters to match the units of the gravitational constant for consistency and accuracy in calculations. Taking the square root of the resulting value yields the escape velocity.

For an intuitive understanding, high escape velocities suggest a strong gravitational pull—common for larger, more massive bodies. Conversely, smaller bodies, like the given asteroid, have a lower escape velocity, indicating a weaker gravitational influence, as seen in the solution with an escape speed of approximately \(263.53 \text{m/s}\).