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  • Supposing Newton’s law of gravitation for gravitation forces F1 and F2 between two masses m1 and m2 at positions r1 and r2 read

Supposing Newton’s law of gravitation for gravitation forces F1 and F2 between two masses m1 and m2 at positions r1 and r2 read

\mathbf{F_{1}}=\mathbf{-F_{2}}=\frac{\mathbf{r_{12}}}{r_{12}^{3}}GM_{0}^{2}\left ( \frac{m_{1}m_{2}}{M_{1}^{2}} \right )^{n} where Mo is a constant of the dimension of mass, \mathbf{r_{12}} = \mathbf{r_{1}} - \mathbf{r_{2}} and n is a number. In such a case,

a) the acceleration due to gravity on earth will be different for different object

b) none of the three laws of Kepler will be valid

c) only the third law will become invalid

d) for n negative, an object lighter than water will sink in water

Answers (1)

The correct answer is the option (a) and (c)

\overrightarrow{F_{1}}=\overrightarrow{F_{2}}=\frac{\overrightarrow{r_{12}}}{r_{12}^{3}}GM_{0}^{2}\left [ \frac{m_{1}m_{2}}{M_{0}^{2}} \right ]^{n}

given, \overrightarrow{r_{12}}=\overrightarrow{r_{1}}-\overrightarrow{r_{2}}

Acceleration due to gravity (g) =

\frac{\left | F \right |}{mass (m)}

g=\left | \frac{GM_{0}^{2}(m_{1}m_{2})^{n}}{r_{12}^{2}M_{0}^{2n}}\times\frac{\hat{r}_{12}}{m} \right |

\hat{r}_{12} tells the direction of g from 1 to 2

g= \frac{GM_{0}^{2}(m_{1}m_{2})^{n}}{r_{12}^{2}M_{0}^{2n}}\; \frac{1}{n}

Since g is dependent on position vector

\overrightarrow{r_{12}}, mass of body m. So, g on earth will be different for different bodies of different mass and their

Position. Hence a and c are correct

As the force is central in nature, Kepler's first and second law is valid rejecting option b.

g= \frac{GM_{0}^{2}(m_{1}m_{2})^{-n}}{r_{12}^{2}M_{0}^{-2n}}\; \frac{1}{n}=\frac{GM_{0}^{2}M_{0}^{2n}}{(m_{1}m_{2})^{n}r_{12}^{2}}\; \frac{1}{m}=\frac{GM_{0}^{2}}{r_{12}^{2}}\left [ \frac{M_{0}^{2}}{m_{1}m_{2}} \right ]\; \frac{1}{m}

g>0

So M_{0}>m_{1}\; or \; m_{2}

The lighter object can sink in water. Hence, d is correct

Posted by

infoexpert23

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