Question 394: Gravitation

Question

A remote-sensing satellite of earth revolves in a circular orbit at a height of 0.25 $$ 	imes $$ 10
6
m above the surface of earth. If earth's radius is 6.38 $$ 	imes $$ 10
6
m and g = 9.8 ms
$$-$$2
, then the orbital speed of the satellite is

Accepted Answer

The orbital speed of the satellite is
$${v_o} = R\sqrt {{g \over {\left( {R + h} ight)}}} $$
where R is the earthÃ¢ÂÂs radius,  g is the acceleration due to gravity on earthÃ¢ÂÂs surface and h is the height above the surface of earth.
Here, R = 6.38 ÃÂ 10
6
m, g = 9.8 m s
Ã¢ÂÂ2
and h = 0.25 ÃÂ 10
6
m
$${v_o} = \left( {6.38 	imes {{10}^6}m} ight)\sqrt {{{9.8\,m{s^{ - 2}}} \over {\left( {6.38 	imes {{10}^6}m + 0.25 	imes {{10}^6}m} ight)}}} $$
= 7.76 ÃÂ 10
3
m s
Ã¢ÂÂ1
= 7.76 km s
Ã¢ÂÂ1

Answer

9.13 km s
$$-$$1

Answer

6.67 km s
$$-$$1

Answer

7.76 km s
$$-$$1

Answer

8.56 km s
$$-$$1

Gravitation

A remote-sensing satellite of earth revolves in a circular orbit at a height of 0.25 $ \times 10 6 m above the surface of earth. If earth's radius is 6.38 \times 10 6 m and g = 9.8 ms -$2 , then the orbital speed of the satellite is

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