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Chapter 6: Problem 705

The acceleration due to gravity near the surface of a planet of radius \(\mathrm{R}\) and density \(\mathrm{d}\) is proportional to (A) \(\mathrm{d} / \mathrm{R}^{2}\) (B) \(\mathrm{d} \mathrm{R}^{2}\) (C) \(\mathrm{dR}\) (D) \(\mathrm{d} / \mathrm{R}\)

Short Answer

Expert verified
The acceleration due to gravity near the surface of a planet of radius R and density d is proportional to dR (option C). This can be found by using Newton's law of gravitation and calculating the mass of the planet using its density and volume. After substituting the mass equation into the acceleration equation and simplifying, we find that g is proportional to dR.

Step by step solution

01

Recall Newton's law of gravitation

The equation for acceleration due to gravity, g, near the surface of a planet is given by Newton's law of gravitation, which is: \(g = \frac{Gm}{R^2}\) where G is the gravitational constant, m is the mass of the planet, and R is its radius.
02

Calculate the mass of the planet

To find the mass of the planet (m), we can use its density (d) and volume (V). The equation for mass is: \(m = dV\) The volume of the sphere (the planet) can be calculated as: \(V = \frac{4}{3}πR^3\) Substituting this expression for V in the mass equation, we get: \(m = d(\frac{4}{3}πR^3)\)
03

Substitute mass equation into acceleration equation

Now, substitute the mass equation into the acceleration due to gravity equation: \(g = \frac{G(d(\frac{4}{3}πR^3))}{R^2}\)
04

Simplify the equation

Simplify the equation by canceling one R from the numerator and denominator: \(g = Gd (\frac{4}{3}πR)\) Now, we can see that g is proportional to dR.
05

Conclusion

Therefore, the acceleration due to gravity near the surface of a planet of radius R and density d is proportional to dR, which corresponds to option (C).

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Most popular questions from this chapter

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