Question

A spherical asteroid has a mass of \(1.769 \cdot 10^{20} \mathrm{~kg} .\) The escape speed from its surface is \(273.7 \mathrm{~m} / \mathrm{s}\). What is the radius of the asteroid?

Short answer

Answer: The radius of the asteroid is approximately \(400,455 \mathrm{m}\).

Step-by-step solution

Write down the given values and the escape speed formula

Given values: Mass of asteroid, \(M = 1.769 \times 10^{20} \mathrm{kg}\) Escape speed, \(v_e = 273.7 \mathrm{m/s}\) Gravitational constant, \(G = 6.674 \times 10^{-11} \mathrm{N}(\mathrm{m}/\mathrm{kg})^2\) Escape speed formula: \(v_e = \sqrt{ \frac{2GM}{r} }\)

Square the escape speed formula and isolate the radius (\(r\))

First, square both sides of the escape speed formula to eliminate the square root: \((v_e)^2 = \frac{2GM}{r}\) Now, we want to isolate \(r\): \(r = \frac{2GM}{(v_e)^2}\)

Plug in the given values and calculate the radius

Now we can plug in the given values for mass, escape speed, and gravitational constant into the formula we derived: \(r = \frac{2 (6.674 \times 10^{-11} \mathrm{N}(\mathrm{m}/\mathrm{kg})^2) (1.769 \times 10^{20} \mathrm{kg})}{(273.7 \mathrm{m/s})^2}\) Calculating the result, we find the radius of the asteroid: \(r \approx 400,455 \mathrm{m}\) So the radius of the asteroid is approximately \(400,455 \mathrm{m}\).