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Chapter 6: Problem 693

If the density of small planet is that of the same as that of the earth while the radius of the planet is \(0.2\) times that of the earth, the gravitational acceleration on the surface of the planet is (A) \(0.2 \mathrm{~g}\) (B) \(0.4 \mathrm{~g}\) (C) \(2 \mathrm{~g}\) (D) \(4 \mathrm{~g}\)

Short Answer

Expert verified
The gravitational acceleration on the surface of the small planet, relative to Earth's gravitational acceleration, is \(g = 0.4 g_e\). Therefore, the correct answer is (B) \(0.4 \mathrm{~g}\).

Step by step solution

01

Write down the formula for gravitational acceleration

The formula for gravitational acceleration on the surface of a planet is given by: \[g = \frac{GM}{R^2}\] where - \(g\) is the gravitational acceleration, - \(G\) is the universal gravitational constant (\(6.674 \times 10^{-11} N(m/kg)^2\)), - \(M\) is the mass of the planet, and - \(R\) is the radius of the planet.
02

Express the mass in terms of density and volume

The mass of the planet can be expressed as: \[M = \rho V\] where - \(\rho\) is the density of the planet, and - \(V\) is the volume of the planet. Since the density is the same for Earth and the small planet, we can represent the volume of the planet as a fraction of Earth's volume: \[V = \frac{4}{3}\pi R^3 = \frac{4}{3}\pi (0.2R_e)^3 = 0.008 \frac{4}{3}\pi R_e^3\] where \(R_e\) is the radius of the Earth. Now we can write the mass of the planet in terms of Earth's mass: \[M = 0.008 M_e\]
03

Substitute the expressions for mass and radius into the formula

Now, we can substitute the expressions for mass and radius of the planet into the gravitational acceleration formula: \[g = \frac{G(0.008 M_e)}{(0.2 R_e)^2}\]
04

Simplify and compare to Earth's gravitational acceleration

We know that Earth's gravitational acceleration, \(g_e\), is given by: \[g_e = \frac{GM_e}{R_e^2}\] Now, divide the gravitational acceleration on the surface of the planet (\(g\)) by the gravitational acceleration on the Earth (\(g_e\)), using the above formulas: \[\frac{g}{g_e} = \frac{\frac{G(0.008 M_e)}{(0.2 R_e)^2}}{\frac{GM_e}{R_e^2}} = \frac{0.008 M_e R_e^2}{0.04 M_e R_e^2} = \frac{2}{5}\]
05

Find gravitational acceleration on the surface of the planet

Finally, we can find the gravitational acceleration on the surface of the planet, relative to Earth's gravitational acceleration: \[g = \frac{2}{5} g_e\] Since Earth's gravitational acceleration, \(g_e\), is approximately 9.81 m/s^2, the gravitational acceleration on the surface of the small planet is approximately: \[g = 0.4 g_e\] The correct answer is (B) \(0.4 \mathrm{~g}\).

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