Two-balls-each-of-radius-R-equal-mass-and-density-are-placed-in-contact-then-the-force-of-gravitation-between-them-is-proportional-to-1-F-1-R-2-2-F-R-3-F-R-4-4-F-1-R-

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Question Number 26959 by Tinkutara last updated on 31/Dec/17

$$\mathrm{Two}\mathrm{balls},\mathrm{each}\mathrm{of}\mathrm{radius}R,\mathrm{equal}\mathrm{mass}$$$$\mathrm{and}\mathrm{density}\mathrm{are}\mathrm{placed}\mathrm{in}\mathrm{contact},\mathrm{then}$$$$\mathrm{the}\mathrm{force}\mathrm{of}\mathrm{gravitation}\mathrm{between}\mathrm{them}$$$$\mathrm{is}\mathrm{proportional}\mathrm{to}$$$$\left(1\right)F\propto \frac{1}{R^2}$$$$\left(2\right)F\propto R$$$$\left(3\right)F\propto R^4$$$$\left(4\right)F\propto \frac{1}{R}$$

Answered by prakash jain last updated on 31/Dec/17

$$\mathrm{mass}\mathrm{of}\mathrm{each}\mathrm{ball}=\frac{4}{3}\rho \pi {\mathrm{R}}^3$$$$\mathrm{F}=\frac{{Gm}_1{m}_2}{R^2}\propto R^4$$$$$$

Commented by Tinkutara last updated on 31/Dec/17

Yes this answer is correct according to one book but I have another book which says:

Commented by Tinkutara last updated on 31/Dec/17

Commented by prakash jain last updated on 31/Dec/17

$$\mathrm{Maybe}ajfour/\mathrm{mrW}1\mathrm{can}\mathrm{comment}.$$$$\mathrm{In}\mathrm{my}\mathrm{view}\mathrm{if}\mathrm{R}\mathrm{is}\mathrm{increased}$$$$\mathrm{mass}\mathrm{will}\mathrm{increase}\mathrm{and}\mathrm{distance}$$$$\mathrm{will}\mathrm{also}\mathrm{increase}\mathrm{so}\mathrm{force}\mathrm{will}$$$$\mathrm{be}\mathrm{proportinal}\mathrm{to}{\mathrm{R}}^4.$$$$\mathrm{In}\mathrm{the}\mathrm{question}\mathrm{density}\mathrm{is}\mathrm{contant}$$$$\mathrm{and}\mathrm{not}\mathrm{mass}.$$

Commented by Tinkutara last updated on 31/Dec/17

But to me both answers seem correct. But what to do if this question really comes in exam?

Commented by Femmy last updated on 31/Dec/17

$$\mathrm{it}\mathrm{is}\mathrm{an}\mathrm{inverse}\mathrm{square}\mathrm{law}\mathrm{remember}?$$

Commented by Tinkutara last updated on 31/Dec/17

But the book from which I posted this question says (3). What is the mistake if we go by other method?

Commented by prakash jain last updated on 31/Dec/17

$${\mathrm{R}}^4\mathrm{is}\mathrm{correct}\mathrm{for}\mathrm{this}\mathrm{question}\mathrm{since}$$$$\mathrm{mass}\mathrm{also}\mathrm{depends}\mathrm{on}R.$$$$F\propto \frac{m_1{m}_2}{r^2}$$$$\mathrm{Given}\rho \mathrm{is}\mathrm{constant}F\alpha R^4$$

Commented by Tinkutara last updated on 31/Dec/17

So when the book from which I posted image is correct?

Commented by mrW1 last updated on 31/Dec/17

$$Ithinktheanswershouldbe\left(1\right).$$$$Thequestionaskstheforcebetween$$$$thetwoballs.Theyarefixunits,$$$$concerningmassandsize.$$

Commented by prakash jain last updated on 31/Dec/17

$$\mathrm{Hi}\mathrm{mrW}1,$$$$\mathrm{The}\mathrm{force}\mathrm{in}\mathrm{this}\mathrm{case}\mathrm{is}\mathrm{proportion}$$$$\mathrm{to}{\mathrm{R}}^4\mathrm{since}\mathrm{mass}\mathrm{is}\mathrm{dependent}\mathrm{on}\mathrm{R}.$$$$\mathrm{Say}\mathrm{another}\mathrm{sumi}\mathrm{lar}\mathrm{question}\mathrm{is}\mathrm{when}$$$$\mathrm{a}\mathrm{ball}\mathrm{is}\mathrm{melted}\mathrm{to}\mathrm{form}\mathrm{a}\mathrm{wire}\mathrm{then}$$$$\mathrm{resistance}\mathrm{is}\mathrm{proportinal}\mathrm{to}{l}^{?}.$$$$\mathrm{Force}\mathrm{between}\mathrm{two}\mathrm{bodies}\mathrm{of}$$$$\mathrm{a}\mathrm{given}\mathrm{mass}\mathrm{is}\mathrm{propotional}\mathrm{to}\frac{1}{R^2}.$$$$\mathrm{But}\mathrm{when}\mathrm{the}\mathrm{question}\mathrm{says}\mathrm{constanst}$$$$\mathrm{density}\left(\mathrm{and}\mathrm{not}\mathrm{given}\mathrm{mass}\right)\mathrm{force}$$$$\mathrm{will}\mathrm{be}\mathrm{propotional}\mathrm{to}{R}^4.$$$$\mathrm{What}\mathrm{is}\mathrm{wrong}\mathrm{with}\mathrm{my}\mathrm{argument}?$$

Commented by mrW1 last updated on 31/Dec/17

$$youarealsorightsir.thequestionis$$$$justnotclearenough.itcanbe$$$$understooddifferently,depending$$$$onifthemassoftheballsisseen$$$$asconstantornot.$$

Answered by Femmy last updated on 31/Dec/17

$$=>$$$$1$$

Commented by Tinkutara last updated on 31/Dec/17

I want to know which option is really correct.

Answered by ajfour last updated on 31/Dec/17

$$massremainsconstantor$$$$densitystaysconstantmustbe$$$$mentioned.$$$$Bothexpeimentsarepossible.$$$$wecanchoosetochangeballs$$$$fornewradiikeepingmassthe$$$$sameordensitythesame.$$

Commented by prakash jain last updated on 31/Dec/17

$$\mathrm{Question}\mathrm{says}\mathrm{copper}\mathrm{balls}\mathrm{so}$$$$\mathrm{density}\mathrm{is}\mathrm{constant}.$$

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