A remote sensing satellite of Earth revolves in a circular orbit at a height of 0 25 × 106 m - Physics - Gravitation

Question

A remote sensing satellite of Earth revolves in a circular orbit at
a height of 0 25 × 106 m above the surface of the
Earth If Earth's radius is 6 38 × 106 m and g = 9
8 m/s2 then the orbital speed of the satellite is
(a) 5 86 km/s (b) 9 13 km/s (c) 6 67 km/s (d) 7 76 km/s

Sayantan · Accepted Answer

Dear Student,

We have,

At equilibrium orbit
Centrifugal Force=Gravitational Forcemv2Ror=GMmRor2or,  v=GMRor   where Ror=Re+h   ; h= distance of the satellite from earth surfacewe also know,g=GMRe2   or,  M=gRe2Gplugging this in eqn for vv=gRe2Re+h⇒7
76 km/s
Regards

A remote sensing satellite of Earth revolves in a circular orbit at a height of 0.25 × 106 m above the surface of the Earth. If Earth's radius is 6.38 × 106 m and g = 9.8 m/s2 then the orbital speed of the satellite is (a) 5.86 km/s (b) 9.13 km/s (c) 6.67 km/s (d) 7.76 km/s

A remote sensing satellite of Earth revolves in a circular orbit at a height of 0.25 × 10⁶ m above the surface of the Earth. If Earth's radius is 6.38 × 10⁶ m and g = 9.8 m/s² then the orbital speed of the satellite is
(a) 5.86 km/s (b) 9.13 km/s (c) 6.67 km/s (d) 7.76 km/s