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Chapter 6: Problem 673

Correct form of gravitational law is (A) $\mathrm{F}=-\left[\left(\mathrm{Gm}_{1} \mathrm{~m}_{2}\right) / \mathrm{r}^{2}\right]$ (B) $\mathrm{F}^{-}=-\left[\left(\mathrm{Gm}_{1} \mathrm{~m}_{2}\right) / \mathrm{r}^{2}\right]$ (C) $\mathrm{F}^{-}=-\left[\left(\mathrm{Gm}_{1} \mathrm{~m}_{2}\right) / \mathrm{r}^{2}\right] \hat{\mathrm{r}}$ (B) $\mathrm{F}^{\rightarrow}=-\left[\left(\mathrm{Gm}_{1} \mathrm{~m}_{2}\right) / \mathrm{r}^{3}\right] \mathrm{r}^{-}$

Short Answer

Expert verified
The correct form of the gravitational law is given by Option C: \(F^{-} = -\frac{(Gm_1m_2)}{r^2}\hat{r}\), as it includes the gravitational constant, the masses of both objects, the distance between their centers, and the force vector \(\hat{r}\) pointing along the line connecting the centers of the objects.

Step by step solution

01

Option A

Option A states: \(F = -\frac{(Gm_1m_2)}{r^2}\) This option does include the gravitational constant and the masses of both objects, and the force goes inversely with the square of the distance between them. However, this option does not include any direction information.
02

Option B

Option B states: \(F^{-} = -\frac{(Gm_1m_2)}{r^2}\) This option is nearly the same as option A, but with a minus sign. The inclusion of a minus sign could indicate the attractive nature of the gravitational force, but this option still does not include any direction information.
03

Option C

Option C states: \(F^{-} = -\frac{(Gm_1m_2)}{r^2}\hat{r}\) This option includes the gravitational constant, the masses of both objects, the distance between their centers, and a unit vector in the radial direction, \(\hat{r}\), which indicates the direction of the force. This seems to be the correct form of the gravitational law.
04

Option D

Option D states: \(F^{\rightarrow} = -\frac{(Gm_1m_2)}{r^3}r^{-} \) This option includes the gravitational constant and the masses of both objects. However, the force is inversely proportional to the cube of the distance, not the square, which is incorrect. Additionally, this option does not have a clear direction. Based on the analysis of all the options, the correct form of the gravitational law is found in Option C.

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Most popular questions from this chapter

A satellite of mass \(\mathrm{m}\) is orbiting close to the surface of the earth (Radius \(\mathrm{R}=6400 \mathrm{~km}\) ) has a K.E. \(\mathrm{K}\). The corresponding \(\mathrm{K} . \mathrm{E}\). of satellite to escape from the earth's gravitational field is (A) \(\mathrm{K}\) (B) \(2 \mathrm{~K}\) (C) \(\mathrm{mg} \mathrm{R}\) (D) \(\mathrm{m} \mathrm{K}\)

At what height over the earth's pole, the free fall acceleration decreases by one percent \(=\ldots \ldots \ldots \mathrm{km}(\mathrm{Re}=6400 \mathrm{~km})\) (A) 32 (B) 80 (C) \(1.253\) (D) 64

Which one of following statements regarding artificial satellite of earth is incorrect (A) The orbital velocity depends on the mass of the satellite (B) A minimum velocity of \(8 \mathrm{kms}^{-1}\) is required by a satellite to orbit quite close to the earth. (C) The period of revolution is large if the radius of its orbit is large (D) The height of geostationary satellite is about \(36000 \mathrm{~km}\) from earth

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) \(24 \sqrt{2}\)

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