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Chapter 8: Problem 32

To what radius would Earth have to shrink, with no change in mass, for escape speed at its surface to be \(30 \mathrm{km} / \mathrm{s} ?\)

Short Answer

Expert verified
The radius to which Earth would have to shrink, with no chance in mass, for escape speed at its surface to be 30 km/s is approximately \(4.57*10^{7} \) meters.

Step by step solution

01

Convert to Appropriate Units

First, convert the given escape speed into meters per second for consistancy with the gravitational constant. This means multiplying 30 km/s by 1000 m/km: \( 30km/s * 1000m/km = 30000 m/s. \)
02

Rearrange Escape Velocity Equation

The formula for escape speed is \( v_{s} = \sqrt {2gr} \). To isolate r, we want to square both sides to get \( v_{s}^{2} = 2gr \) then divide by \( 2g \) to get \( r = v_{s}^{2} / 2g \)
03

Substitute known values

We know the escape speed we want (\( v_{s} = 30000 m/s \)), and we know the acceleration due to gravity (\( g = 9.81 m/s^{2} \)). Substitute these into equation to get \( r = ( 30000 m/s )^{2} / 2* (9.81 m/s^{2}) \)
04

Calculate R

After doing the computation, we'll get \( r \approx 4.57*10^{7}m \)

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Most popular questions from this chapter

When you stand on Earth, the distance between you and Earth is zero. So why isn't the gravitational force infinite?

A projectile is launched vertically upward from a planet of mass \(M\) and radius \(R ;\) its initial speed is \(\sqrt{2}\) times the escape speed. Derive an expression for its speed as a function of the distance \(r\) from the planet's center.

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To what fraction of its current radius would Earth have to shrink (with no change in mass) for the gravitational acceleration at its surface to triple?

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