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AfilCa [17]
3 years ago
13

4. A 1,750 kg weather satellite moves in a circular orbit with a gravitational

Chemistry
1 answer:
Zolol [24]3 years ago
7 0

Explanation:

It is known that the value of free fall acceleration is g = 6.44 m/s^{2}. And, the value of radius of Earth is 6.4 \times 10^{6} m.

Let us assume that height of the satellite is h.

It is known that,

             g = \frac{GM}{r^{2}}        

where,   r = (R + h)

Hence,  

              r = \sqrt{\frac{GM}{g}}

              r = \sqrt{\frac{6.67 \times 10^{-11} \times 5.98 \times 10^{24}}{6.44 m/s}}

                = 7.9 \times 10^{6} m

Now, formula for height of satellite above the Earth's surface is as follows.

            h = r - R

               = 7.9 \times 10^{6} - 6.4 \times 10^{6}

               = 1.5 \times 10^{6} m

Thus, we can conclude that the satellite is 1.5 \times 10^{6} m high above Earth's surface.

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