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acceleration due to gravity will be (i) one-half, (ii) one-fourth of the acceleration due to gravity at the surface of the earth ? Radius of the earth is sidereal revolution periods of the planets Vent $6400\mathrm{km}$. Ans. (i) $3200\mathrm{km}$, (ii) $4800\mathrm{km}$. Estimate the mean distances of these planets fro 12. The radius of the earth is $6.4\times 10^6\mathrm{m}$ and the 15. The mass and radius of earth are $6.0\times 10^{24}$ mean density is $5.5\times 10^3\mathrm{kg}\mathrm{m}$. Determine $6400\mathrm{km}$. What will be the potential energ the gravitational potential at the earth's surface. $200\mathrm{kg}$ body placed $600\mathrm{km}$ high from earth's s $\left(G=6.67\times 10^{-11}\mathrm{N}{\mathrm{m}}^2{\mathrm{kg}}^{-2}\right)$ Ans. $-6.3\times 10^7\mathrm{J}{\mathrm{kg}}^{-1}.(G^{\prime }=6.7\times 10^{-11}$ SI units). Ans. $-1.15\times$ 13. Two bodies of masses $100\mathrm{kg}$ and $1000\mathrm{kg}$ are at a 16. How much energy would be needed for a 100 distance $1.00\mathrm{m}$ apart. Calculate the gravitational field intensity and the potential at the middle-point of the line joining them. $\left(G=6.67\times 10^{-11}\mathrm{N}{\mathrm{m}}^2{\mathrm{kg}}^{-2}\right)$. Ans. $2.40\times 10^{-7}\mathrm{N}{\mathrm{kg}}^{-1},-1.47\times 10^{-7}\mathrm{J}{\mathrm{kg}}^{-1}$. to escape from the earth ? $(g=10\mathrm{m}{\mathrm{s}}^{-2}$ and $\mathrm{I}$ the earth $R_e=6.4\times 10^6\mathrm{m}$.) Hint : Energy $=G{M}_{e}m/R_e=mg{R}_e,\therefore GM$ Ans. $6.4$ 14. The mass of the earth is $6.0\times 10^{24}\mathrm{kg}$. Calculate, 17. A $500-\mathrm{kg}$ artificial satellite is revolving arc with sign, (i) the potential energy of a body of mass earth at a height of $1800\mathrm{km}$ from the earth. $33.5\mathrm{kg}$ and (ii) the gravitational potential at a the potential energy, kinetic energy and tot distance of $3.35\times 10^{10}\mathrm{m}$ from the centre of the earth $(G=6.7\times 10^{-11}$ SI units). Ans. (i) $-4.02\times 10^5\mathrm{J}$, (ii) $-12\times 10^3\mathrm{J}{\mathrm{kg}}^{-1}$. of the satellite. The earth's radius is 6400 $g=10\mathrm{m}{\mathrm{s}}^{-2}$. III Based on Satellites and Escape Velocity 18. The distance of the moon from the earth is Ans. (i) $7.36\times 10^3\mathrm{m}{\mathrm{s}}^{-1}$, (ii) $9.95\times 10$ $3.8\times 10^5\mathrm{km}$. Calculate the speed of the moon revolving around the earth. Mass of the earth 22. A satellite is revolving around the earth in $=6.1\times 10^{24}\mathrm{kg}$ and $G=6.7\times 10^{-11}\mathrm{N}{\mathrm{m}}^2{\mathrm{kg}}^{-2}$. orbit of radius $8000\mathrm{km}$. With what speed

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Updated on: Dec 4, 2022