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Chapter 6: Problem 40

The escape speed from the surface of the Earth is $11.2 \mathrm{km} / \mathrm{s} .$ What would be the escape speed from another planet of the same density (mass per unit volume) as Earth but with a radius twice that of Earth?

Short Answer

Expert verified
Answer: The escape speed from the surface of the other planet is 22.4 km/s.

Step by step solution

01

Understand the escape speed formula

The escape speed formula is given by: $$ v_e = \sqrt{ \frac{2GM}{R}} $$ Where \(v_e\) is the escape speed, \(G\) is the gravitational constant, \(M\) is the mass of the planet, and \(R\) is the distance from the center of mass (in this case, the radius of the planet).
02

Find the mass of the other planet

Since the density of the other planet is the same as Earth and its radius is twice that of Earth, we can find the mass of the other planet using the formula: $$ M_m = \rho V_m = \rho \frac{4}{3}\pi R_m^3 $$ Where \(M_m\) is the mass of the other planet, \(\rho\) is the density, \(V_m\) is the volume of the other planet, and \(R_m\) is the radius of the other planet, which is 2 times the Earthradius, \(2R_e\). We also know the mass of Earth in terms of its density: $$ M_e = \rho V_e = \rho \frac{4}{3}\pi R_e^3 $$
03

Determine the ratio of the planet's mass to Earth's mass

We need to find the ratio of the planet's mass to Earth's mass: $$ \frac{M_m}{M_e} = \frac{\rho \frac{4}{3}\pi (2R_e)^3}{\rho \frac{4}{3}\pi R_e^3} = \frac{8R_e^3}{R_e^3} = 8 $$ So we have now that the mass of the other planet, \(M_m\), is 8 times the mass of Earth, \(M_e\).
04

Calculate the escape speed of the other planet

Using the escape speed formula, the escape speed of the other planet is: $$ v_{em} = \sqrt{ \frac{2G \cdot 8 M_e }{2R_e}} $$ Now we plug in the given escape speed of Earth: $$ 11.2 \mathrm {km/s} = \sqrt{ \frac{2GM_e}{R_e}} $$ Squaring both sides gives: $$ 11.2^2 = \frac{2GM_e}{R_e} $$ We will use this value to solve for the escape speed of the other planet, \(v_{em}\): $$ v_{em} = \sqrt{ \frac{2G \cdot 8 M_e }{2R_e}} = \sqrt{ \frac{8 \cdot 11.2^2 }{2}}=11.2\sqrt{ \frac{8}{2}} $$ And finally, we have: $$ v_{em} = 11.2\sqrt{4} = 11.2 \cdot 2 = 22.4 \mathrm{km/s} $$
05

Final answer

The escape speed from the surface of the other planet is \(22.4 \mathrm{km/s}\).

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