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Q9 P

Expert-verifiedFound in: Page 439

Book edition
9th Edition

Author(s)
Raymond A. Serway, John W. Jewett

Pages
1624 pages

ISBN
9781133947271

**Two objects attract each other with a gravitational force of a magnitude**** $1.00\times {10}^{-8}\text{N}$when separated by 20.0 cm. If the total mass of the two objects is 5.00 kg, what is the mass of each?**

The masses of objects are ${m}_{1}=3.00kg$and ${m}_{2}=2.00kg$ .

Newton’s law of universal gravitation states that the gravitational force of attraction between any two particles of masses ${M}_{1}$and ${M}_{2}$ separated by a distance *r* has the magnitude:

${F}_{g}=G\frac{{m}_{1}.{m}_{2}}{{r}^{2}}$

Where, $G=6.674\times {10}^{-11}\text{N}\xb7{\text{m}}^{\text{2}}{\text{/kg}}^{\text{2}}$ is the universal gravitational constant. This equation enables us to calculate the force of attraction between masses under many circumstances.

Gravitational force: ${F}_{g}=1.00\times {10}^{-8}\text{N}$

Distance between two objects: $r=20\text{cm}=0.2\text{m}$

We find the reason for the above question by finding the mass of sphere.

Let suppose the first mass ${m}_{1},$is and the other mass will be ${m}_{2}=5-{m}_{1},$

According to Newton’s law of universal gravitation, we have

${F}_{g}=G\frac{{m}_{1}.{m}_{2}}{{r}^{2}}\phantom{\rule{0ex}{0ex}}1\times {10}^{-8}=6.67\times {10}^{-11}\times \frac{{m}_{1}.\left(5-{m}_{1}\right)}{{0.2}^{2}}..............\left(\because {m}_{2}=5-{m}_{1}\right)\phantom{\rule{0ex}{0ex}}{m}_{1}.\left(5-{m}_{1}\right)=\frac{0.04\times {10}^{-8}}{{10}^{-11}\times 6.67}\phantom{\rule{0ex}{0ex}}{m}_{1}.\left(5-{m}_{1}\right)=5.99\approx 6\phantom{\rule{0ex}{0ex}}{{m}_{1}}^{2}-5{m}_{1}+6=0$

On simplifying and solving, we get,

${m}_{1}=3.00\text{kg}\phantom{\rule{0ex}{0ex}}{m}_{2}=5-{m}_{1}=5-3.00=2.00\text{kg}$

Hence, the masses of objects are ${m}_{1}=3.00kg$ and ${m}_{2}=2.00kg$.

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