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Question Number 27283    Answers: 0   Comments: 0

Question Number 27074    Answers: 2   Comments: 0

Question Number 26959    Answers: 3   Comments: 0

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)

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

Question Number 25684    Answers: 1   Comments: 0

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) . If the satellite above is synchronous how fast is it moving through space,taking the period to be 24hrs and M_e =mass of the satellite and is equal to 5.98×10^(24) kg

$${For}\:{a}\:{geosynchronous}\:{satellite}\:{of} \\ $$$${mass}\:{m}\:{moving}\:{in}\:{a}\:{circular} \\ $$$${orbit}\:{around}\:{the}\:{earth}\:{at}\:{a}\:{constant} \\ $$$${speed}\:{v}\:{and}\:{an}\:{altitude}\:{h}\:{above} \\ $$$${the}\:{earth}\:{surface}.{Show}\:{the} \\ $$$${velocity}\:{v}=\left(\frac{{GM}_{{e}} }{{R}_{{e}} +{h}}\right)^{\frac{\mathrm{1}}{\mathrm{2}}} . \\ $$$$ \\ $$$${If}\:{the}\:{satellite}\:{above}\:{is}\:{synchronous} \\ $$$${how}\:{fast}\:{is}\:{it}\:{moving}\:{through} \\ $$$${space},{taking}\:{the}\:{period}\:{to}\:{be}\:\mathrm{24}{hrs} \\ $$$${and}\:{M}_{{e}} ={mass}\:{of}\:{the}\:{satellite}\:{and} \\ $$$${is}\:{equal}\:{to}\:\mathrm{5}.\mathrm{98}×\mathrm{10}^{\mathrm{24}} {kg} \\ $$

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