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WITCHER [35]
2 years ago
13

A 4500-kg spaceship is in a circular orbit 190 km above the surface of Earth. It needs to be moved into a higher circular orbit

of 380 km to link up with the space station at that altitude. In this problem you can take the mass of the Earth to be 5.97 × 1024 kg.
I am confused how to solve this problem, please show how this answer is obtained.
Physics
1 answer:
Maslowich2 years ago
8 0

Answer:

Explanation:

Total energy of a satellite in an orbit , h height away

= -  GMm /2 ( R + h )

When h = 380 km

Total energy of a satellite = \frac{6.67\times10^{-11}\times5.97\times10^{24}\times 4500}{2\times(6378+380)\times10^3}

=  - 13.25 x 10¹⁰ J

When h = 190 km

Total energy of a satellite =

\frac{6.67\times10^{-11}\times5.97\times10^{24}\times 4500}{2\times(6378+190)\times10^3}

=   - 13.63 x 10¹⁰ J

Diff

= 38 x 10⁸ J Energy will be required.

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Taking out ‘mg’ as common and we get

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If in case, the meter stick is to be at balanced form, then the acting torques sum would be zero. So,

                   4 m g(x)+(m g)(x-50 c m)=0

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Taking out ‘mg’ as common and we get

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                   x=\frac{50}{5}=10 \mathrm{cm}

Hence, the stick should be pivoted at a distance of,

                   x^{\prime}=100 \mathrm{cm}-10 \mathrm{cm}=10 \mathrm{cm}

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