Question

What is the ratio of the escape speed to the orbital speed of a satellite at the surface of the Moon, where the gravitational acceleration is about a sixth of that on Earth?

Short answer

Answer: The ratio of the escape speed to the orbital speed of a satellite at the surface of the Moon is approximately 1.414.

Step-by-step solution

Write down the given information

The gravitational acceleration on the Moon is a sixth of that on Earth. g_moon = (1/6) * g_earth

Find the formulas for escape speed and orbital speed

For escape speed, we have: v_escape = sqrt(2 * G * M / R) For orbital speed, we have: v_orbital = sqrt(G * M / R) where: v_escape = escape speed v_orbital = orbital speed G = gravitational constant M = mass of the celestial body (in this case, the Moon) R = radius of the celestial body

Calculate the ratio of escape speed to orbital speed

To find the ratio of escape speed to orbital speed, we will divide the escape speed formula by the orbital speed formula: ratio = v_escape / v_orbital ratio = (sqrt(2 * G * M / R)) / (sqrt(G * M / R))

Simplify the ratio

The gravitational constant (G) and the mass of the Moon (M) will cancel out in the ratio, so we are left with: ratio = sqrt(2 * R / R) ratio = sqrt(2)

State the final result

The ratio of the escape speed to the orbital speed of a satellite at the surface of the Moon is: ratio = sqrt(2) ≈ 1.414 So the escape speed is about 1.414 times greater than the orbital speed.