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Category: Gravitation

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

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For-a-geosynchronous-satellite-of-mass-m-moving-in-a-circular-orbit-around-the-earth-at-a-constant-speed-v-and-an-altitude-h-above-the-earth-surface-Show-the-velocity-v-GM-e-R-e-h-1-2-I

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Question Number 25684 by NECx last updated on 13/Dec/17 $${For}\:{a}\:{geosynchronous}\:{satellite}\:{of} \\ $$$${mass}\:{m}\:{moving}\:{in}\:{a}\:{circular} \\ $$$${orbit}\:{around}\:{the}\:{earth}\:{at}\:{a}\:{constant} \\ $$$${speed}\:{v}\:{and}\:{an}\:{altitude}\:{h}\:{above} \\ $$$${the}\:{earth}\:{surface}.{Show}\:{the} \\ $$$${velocity}\:{v}=\left(\frac{{GM}_{{e}} }{{R}_{{e}} +{h}}\right)^{\frac{\mathrm{1}}{\mathrm{2}}} . \\ $$$$…

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A-rocket-vertically-from-the-surface-of-the-earth-with-an-initil-velocity-v-o-show-that-its-velocity-v-at-height-h-is-given-by-v-o-2-v-2-2gh-1-h-R-where-R-is-radius-of-earth-and-g-is-the

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Question Number 74411 by peter frank last updated on 24/Nov/19 $${A}\:{rocket}\:{vertically}\:{from} \\ $$$${the}\:{surface}\:{of}\:{the}\:{earth} \\ $$$${with}\:{an}\:{initil}\:{velocity}\left({v}_{{o}} \right) \\ $$$${show}\:{that}\:{its}\:{velocity}\:{v} \\ $$$${at}\:{height}\:{h}\:{is}\:{given}\:{by} \\ $$$${v}_{{o}} ^{\mathrm{2}} −{v}^{\mathrm{2}} =\frac{\mathrm{2}{gh}}{\mathrm{1}+\frac{{h}}{{R}}}…

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Question-138896

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Question Number 138896 by ajfour last updated on 19/Apr/21 Answered by ajfour last updated on 20/Apr/21 $${Let}\:{A}\:{be}\:{origin}\:{and}\:{x}\:{axis} \\ $$$${vertically}\:{down}\:{towards}\:{center} \\ $$$${of}\:{cube}. \\ $$$$−{dg}_{{x}} =\frac{\left({Gdm}\right){x}}{\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}}…

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