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Chapter 8: Problem 15

The mass of a planet is double the mass of the earth and its radius is half of that of the earth. If \(\mathrm{F}\) is the force of attraction on an object on the surface of earth, then the force of attraction on the same object on the surface of that planet is (1) \(2 \mathrm{~F}\) (2) \(4 \mathrm{~F}\) (3) \(\underline{\mathrm{F}}\) (4) \(8 \mathrm{~F}\)

Short Answer

Expert verified
Answer: The gravitational force of attraction on the object on the surface of the other planet is 8 times the force on that object on the surface of Earth.

Step by step solution

01

Write down the formula for gravitational force

The formula for gravitational force between two objects is given by: \(\mathrm{F} = G \frac{m_{1}m_{2}}{r^{2}}\) where \(\mathrm{F}\) is the gravitational force, \(G\) is the gravitational constant (\(6.674 \times 10^{-11} ~\mathrm{Nm^2/kg^2}\)), \(m_{1}\) and \(m_{2}\) are the masses of the objects, and \(r\) is the distance between the centers of the two objects. In our case, since we are calculating the force of attraction on the surface of the planets, \(r\) will be equal to the radius of the respective planet.
02

Apply the formula for gravitational force on Earth

For Earth, let \(F_{\rm E}\) be the force of attraction, \(M_{\rm E}\) be the mass of Earth, and \(R_{\rm E}\) be the radius of Earth. Using the gravitational force formula: \(F_{\rm E} = G \frac{mM_{\rm E}}{R_{\rm E}^2}\)
03

Apply the formula for the other planet

For the other planet, let \(F_{\rm P}\) be the force of attraction, \(M_{\rm P} = 2M_{\rm E}\) be its mass (double the Earth's mass), and \(R_{\rm P} = 0.5R_{\rm E}\) be its radius (half the Earth's radius). Using the gravitational force formula: \(F_{\rm P} = G \frac{m(2M_{\rm E})}{(0.5R_{\rm E})^2}\)
04

Compare the two forces of attraction

To find the force of attraction on the object on the surface of the other planet, we will compare \(F_{\rm P}\) to \(F_{\rm E}\). Divide both sides of the \(F_{\rm P}\) equation by \(F_{\rm E}\): \(\frac{F_{\rm P}}{F_{\rm E}} = \frac{G \frac{m(2M_{\rm E})}{(0.5R_{\rm E})^2}}{G \frac{mM_{\rm E}}{R_{\rm E}^2}}\) Now, cancel out \(G\), \(m\) and \(M_{\rm E}\): \(\frac{F_{\rm P}}{F_{\rm E}} = \frac{2}{(0.5)^2} = \frac{2}{0.25} = 8\)
05

Find the force of attraction on the other planet

Since \(\frac{F_{\rm P}}{F_{\rm E}} = 8\), we find that the force of attraction on the object on the surface of the other planet is \(8F_{\rm E}\). So, the correct answer is (4) \(8 \mathrm{~F}\).

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Most popular questions from this chapter

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Two identical solid spheres each of radius 'r' are in contact with each other. If the gravitati between them is \(\mathrm{F}\), then which of the following relations is correct? (1) \(\mathrm{F} \propto \frac{1}{\mathrm{r}}\) (2) \(\mathrm{F} \propto \frac{1}{\mathrm{r}^{2}}\) (3) \(\mathrm{F} \propto \frac{1}{\mathrm{r}^{3}}\) (4) \(\mathrm{F} \propto \frac{1}{\mathrm{r}^{4}}\)

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