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प्रश्न
Calculate the critical velocity of the satellite to be located at 35780 km above the surface of earth.
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उत्तर
Given: Height of the satellite above the earth’s surface (h)
= 35780 km
= 35780 × 103 m
We know that:
Gravitational constant (G) = 6.67 × 10–11 N m2/kg2,
mass of earth (M) = 6 × 1024 kg,
radius of earth (R) = 6400 km
= 6400 × 103 m
To find: Tangential velocity of satellite (vc)
Formula: `"v"_"c" = sqrt("GM"/("R + h"))`
Calculation: From formula,
`"v"_"c" = sqrt(((6.67 xx 10^-11) xx (6 xx 10^24))/((6400 + 35780) xx 10^3))`
`= sqrt((40.02 xx 10^13)/(42180 xx 10^3))`
`= sqrt(40.02/42180 xx 10^10)`
`= sqrt (0.0009487909 xx 10^10)`
`= sqrt(9.487909 xx10^6)`
≈ `sqrt9.5 xx 10^3`
= 3.08 × 103 m/s
= 3.08 km/s
The critical velocity of the satellite is 3.08 km/s.
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संबंधित प्रश्न
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| Earth’s surface (average) | 8.8 | 0.225 |
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The orbit of a satellite is exactly 35780 km above the earth's surface and its tangential velocity is 3.08 km/s.
How much time the satellite will take to complete one revolution around the earth?
(Radius of earth = 6400 km.)
