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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

A body weights \(700 \mathrm{~g} \mathrm{wt}\) on the surface of earth. How much it weight on the surface of planet whose mass is \(1 / 7\) and radius is half that of the earth (A) \(200 \mathrm{~g} \mathrm{wt}\) (B) \(400 \mathrm{~g} \mathrm{wt}\) (C) \(50 \mathrm{~g} \mathrm{wt}\) (D) \(300 \mathrm{~g}\) wt.

The largest and shortest distance of the earth from the sun are \(\mathrm{r}_{1}\) and \(\mathrm{r}_{2}\) its distance from the sun when it is at the perpendicular to the major axis of the orbit drawn from the sun (A) \(\left[\left(\mathrm{r}_{1}+\mathrm{r}_{2}\right) / 4\right]\) (B) $\left[\left(\mathrm{r}_{1} \mathrm{r}_{2}\right) /\left(\mathrm{r}_{1}+\mathrm{r}_{2}\right)\right]$ (C) \(\left[\left(2 r_{1} r_{2}\right) /\left(r_{1}+r_{2}\right)\right]\) (D) \(\left[\left(\mathrm{r}_{1}+\mathrm{r}_{2}\right) / 3\right]\)

3 particle each of mass \(\mathrm{m}\) are kept at vertices of an equilateral triangle of side \(L\). The gravitational field at center due to these particles is (A) zero (B) \(\left[(3 \mathrm{GM}) / \mathrm{L}^{2}\right]\) (C) \(\left[(9 \mathrm{GM}) / \mathrm{L}^{2}\right]\) (D) \((12 / \sqrt{3})\left(\mathrm{Gm} / \mathrm{L}^{2}\right)\)

4 A satellite which is geostationary in a particular orbit is taken to another orbit. Its distance from the center of earth in new orbit is two times of the earlier orbit. The time period in second orbit is $\ldots \ldots \ldots \ldots$ hours. (A) \(4.8\) (B) \(48 \sqrt{2}\) (C) 24 (D) \(24 \sqrt{2}\)

At what distance from the center of earth, the value of acceleration due to gravity \(g\) will be half that of the surfaces \((\mathrm{R}=\) Radius of earth \()\) (A) \(2 \mathrm{R}\) (B) \(\mathrm{R}\) (C) \(1.414 \mathrm{R}\) (D) \(0.414 \mathrm{R}\)
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