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NEET-XII-Physics

11: Gravitation

with Solutions - page 7
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  • #32
    A satellite of mass 1000 kg is supposed to orbit the earth at a height of 2000 km above the earth’s surface. Find (a) its speed in the orbit, (b) is kinetic energy, (c) the potential energy of the earth-satellite system and (d) its time period. Mass of the earth = 6 × 1024 kg. (a) its speed in the orbit, (b) is kinetic energy, (c) the potential energy of the earth-satellite system and (d) its time period. Mass of the earth = 6 × 1024 kg.
    Ans : (a) Speed of the satellite in its orbit
    `` v=\sqrt{\frac{\,\mathrm{\,G\,}M}{r+h}}=\sqrt{\frac{g{r}^{2}}{r+h}}``
    `` \Rightarrow v=\sqrt{\frac{9.8\times {\left(6400\times {10}^{3}\right)}^{2}}{{10}^{6}\times \left(6.4+2\right)}}``
    `` \Rightarrow v=\sqrt{\frac{9.8\times 6.4\times 6.4\times {10}^{6}}{8.4}}``
    `` \Rightarrow v=6.9\times {10}^{3}\,\mathrm{\,m\,}/\,\mathrm{\,s\,}=6.9\,\mathrm{\,km\,}/\,\mathrm{\,s\,}`` (b) Kinetic energy of the satellite
    `` \,\mathrm{\,K\,}.\,\mathrm{\,E\,}.=\frac{1}{2}m{v}^{2}``
    `` =\frac{1}{2}\times 1000\times {\left(6.9\times {10}^{3}\right)}^{2}``
    `` =\frac{1}{2}\times 1000\times \left(47.6\times {10}^{6}\right)``
    `` =2.38\times {10}^{10}\,\mathrm{\,J\,}`` (c) Potential energy of the satellite
    `` \,\mathrm{\,P\,}.\,\mathrm{\,E\,}.=-\frac{\,\mathrm{\,GM\,}m}{\left(\,\mathrm{\,R\,}+h\right)}``
    `` =-\frac{6.67\times {10}^{-11}\times 6\times {10}^{24}\times {10}^{3}}{\left(6400+2000\right)\times {10}^{3}}``
    `` =\frac{40\times {10}^{13}}{8400}=-4.76\times {10}^{10}\,\mathrm{\,J\,}`` (d) Time period of the satellite
    `` T=\frac{2\,\mathrm{\,\pi \,}\left(r+h\right)}{\,\mathrm{\,v\,}}``
    `` =\frac{2\times 3.14\times 8400\times {10}^{3}}{6.9\times {10}^{3}}``
    `` =\frac{6.28\times 84\times {10}^{2}}{6.9}``
    `` =76.6\times 10.2\,\mathrm{\,s\,}``
    `` =2.1\,\mathrm{\,h\,}``
    Page No 227: (a) Speed of the satellite in its orbit
    `` v=\sqrt{\frac{\,\mathrm{\,G\,}M}{r+h}}=\sqrt{\frac{g{r}^{2}}{r+h}}``
    `` \Rightarrow v=\sqrt{\frac{9.8\times {\left(6400\times {10}^{3}\right)}^{2}}{{10}^{6}\times \left(6.4+2\right)}}``
    `` \Rightarrow v=\sqrt{\frac{9.8\times 6.4\times 6.4\times {10}^{6}}{8.4}}``
    `` \Rightarrow v=6.9\times {10}^{3}\,\mathrm{\,m\,}/\,\mathrm{\,s\,}=6.9\,\mathrm{\,km\,}/\,\mathrm{\,s\,}`` (b) Kinetic energy of the satellite
    `` \,\mathrm{\,K\,}.\,\mathrm{\,E\,}.=\frac{1}{2}m{v}^{2}``
    `` =\frac{1}{2}\times 1000\times {\left(6.9\times {10}^{3}\right)}^{2}``
    `` =\frac{1}{2}\times 1000\times \left(47.6\times {10}^{6}\right)``
    `` =2.38\times {10}^{10}\,\mathrm{\,J\,}`` (c) Potential energy of the satellite
    `` \,\mathrm{\,P\,}.\,\mathrm{\,E\,}.=-\frac{\,\mathrm{\,GM\,}m}{\left(\,\mathrm{\,R\,}+h\right)}``
    `` =-\frac{6.67\times {10}^{-11}\times 6\times {10}^{24}\times {10}^{3}}{\left(6400+2000\right)\times {10}^{3}}``
    `` =\frac{40\times {10}^{13}}{8400}=-4.76\times {10}^{10}\,\mathrm{\,J\,}`` (d) Time period of the satellite
    `` T=\frac{2\,\mathrm{\,\pi \,}\left(r+h\right)}{\,\mathrm{\,v\,}}``
    `` =\frac{2\times 3.14\times 8400\times {10}^{3}}{6.9\times {10}^{3}}``
    `` =\frac{6.28\times 84\times {10}^{2}}{6.9}``
    `` =76.6\times 10.2\,\mathrm{\,s\,}``
    `` =2.1\,\mathrm{\,h\,}``
    Page No 227:
  • #32-a
    its speed in the orbit,
    Ans : Speed of the satellite in its orbit
    `` v=\sqrt{\frac{\,\mathrm{\,G\,}M}{r+h}}=\sqrt{\frac{g{r}^{2}}{r+h}}``
    `` \Rightarrow v=\sqrt{\frac{9.8\times {\left(6400\times {10}^{3}\right)}^{2}}{{10}^{6}\times \left(6.4+2\right)}}``
    `` \Rightarrow v=\sqrt{\frac{9.8\times 6.4\times 6.4\times {10}^{6}}{8.4}}``
    `` \Rightarrow v=6.9\times {10}^{3}\,\mathrm{\,m\,}/\,\mathrm{\,s\,}=6.9\,\mathrm{\,km\,}/\,\mathrm{\,s\,}``
  • #32-b
    is kinetic energy,
    Ans : Kinetic energy of the satellite
    `` \,\mathrm{\,K\,}.\,\mathrm{\,E\,}.=\frac{1}{2}m{v}^{2}``
    `` =\frac{1}{2}\times 1000\times {\left(6.9\times {10}^{3}\right)}^{2}``
    `` =\frac{1}{2}\times 1000\times \left(47.6\times {10}^{6}\right)``
    `` =2.38\times {10}^{10}\,\mathrm{\,J\,}``
  • #32-c
    the potential energy of the earth-satellite system and
    Ans : Potential energy of the satellite
    `` \,\mathrm{\,P\,}.\,\mathrm{\,E\,}.=-\frac{\,\mathrm{\,GM\,}m}{\left(\,\mathrm{\,R\,}+h\right)}``
    `` =-\frac{6.67\times {10}^{-11}\times 6\times {10}^{24}\times {10}^{3}}{\left(6400+2000\right)\times {10}^{3}}``
    `` =\frac{40\times {10}^{13}}{8400}=-4.76\times {10}^{10}\,\mathrm{\,J\,}``
  • #32-d
    its time period. Mass of the earth = 6 × 1024 kg.
    Ans : Time period of the satellite
    `` T=\frac{2\,\mathrm{\,\pi \,}\left(r+h\right)}{\,\mathrm{\,v\,}}``
    `` =\frac{2\times 3.14\times 8400\times {10}^{3}}{6.9\times {10}^{3}}``
    `` =\frac{6.28\times 84\times {10}^{2}}{6.9}``
    `` =76.6\times 10.2\,\mathrm{\,s\,}``
    `` =2.1\,\mathrm{\,h\,}``
    Page No 227: