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Pratham

  • 2 years ago

An artificial satellite of mass M revolves in circular orbit whose radius is 4R where R is radius of earth. there is dust in bet. 2R & 4R .if mass of satellite is increasing at rate of u per sec. find time in which satellite come to orbit of radius 2R.

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  1. Jemurray3
    • 2 years ago
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    The radius of the orbit is independent of mass but is dependent on velocity, so you need to find how the velocity changes as the thing picks up mass. Since momentum is constant, we can write that \[v = \frac{p_0}{m} \] where \[p_0 = Mv_0 \] Equating the gravitational attraction with the necessary centripetal force, \[ G\frac{m_{earth}m_{satellite}}{r^2} = \frac{m_{satellite}v^2}{r}\] or \[ r = \frac{Gm_{earth}}{v^2} \] plugging in the earlier equation, \[r = \frac{Gm_{earth}m^2_{satellite}}{M_0^2v_0^2} \] Lastly, you can write the mass of the satellite as a function of time: \[m_{satellite} = M_0 + u\cdot t \] where u is the rate of increase of the mass. Solve for t.

  2. Mikael
    • 2 years ago
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    I suppose this is solvable without analyzing the transition process. Only the final and the initial potential and kinetic energies matter. And these are functions of Radius-s and Mass-s at the beginning and the end of the process.

  3. Pratham
    • 2 years ago
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    jemurray3,can you solve the problem thoroughly cuse i still dont get it

  4. Jemurray3
    • 2 years ago
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    The above equation gives the radius of the orbit as a function of the mass of the satellite. You can invert it: \[ m_{satellite} = \sqrt{\frac{M_0^2v_0^2}{G m_{earth}}\cdot r} \] And since the mass of the satellite increases at the rate of u kg/s (I assume) \[m_{satellite} = M_0 + u\cdot t = \sqrt{\frac{M_0^2v_0^2}{G m_{earth}}\cdot r} \] which means that \[ t = \frac{1}{u} \left[ \sqrt{\frac{M_0^2v_0^2}{G m_{earth}}\cdot r} - M_0\right] \] where u is the rate of mass increase, M_0 is the initial mass of the satellite, v_0 is the initial mass of the satellite, and r is the desired radius. This could also be restated, since \[ \frac{M_0v_0^2}{r_i} = \frac{Gm_{earth} M_0 }{r_i^2} \implies v_0^2 = \frac{Gm_{earth}}{r_i} \] so we can also say that \[t = \frac{1}{u} \left[ \sqrt{ \frac{M_0^2 r_f}{r_i}} - M_0 \right] = \frac{M_0}{u} \left[ \sqrt{\frac{r_f}{r_i}}-1\right] \] Which is much simpler.

  5. Pratham
    • 2 years ago
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    is there no effect of dust present?

  6. Jemurray3
    • 2 years ago
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    The mass increase is due to the pile-up of dust on the satellite.

  7. Pratham
    • 2 years ago
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    oh right,thank you very much!!!!!!!!

  8. Jemurray3
    • 2 years ago
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    Hold on... there's a typo in there somewhere, let me find it...

  9. Jemurray3
    • 2 years ago
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    Nevermind, there isn't. I don't think. But I should point out that this equation means that the radius should increase over time, rather than decrease, which is a result of the assumption of inelastic collisions between the dust and the satellite.

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