Question

A synchronous satellite goes around the earth once in every 24 h. What is the radius of orbit of the synchronous satellite in terms of the earth’s radius ?

(Given : Mass of the earth,

radius of the earth, $R _ { E } = 6.37 \times 10 ^ { 6 }$ m, universal constant of gravitation, $\mathrm { G } = 6.67 \times 10 ^ { - 11 } \mathrm {~N} \mathrm {~m} ^ { 2 } \mathrm {~kg} ^ { - 2 }$

• A. $2.4 \mathrm { R } _ { \mathrm { E } }$
• B. $3.6 R _ { E }$
• C.
• D. $6.6 R _ { E }$

Answer 1

Answer: (D)

$6.6 R _ { E }$

Answer 2

The time period of satellite is

$\mathrm { T } = 2 \pi \sqrt { \frac { \mathrm { r } ^ { 3 } } { \mathrm { GM } _ { \mathrm { E } } } }$

Squaring both sides, we get

$\mathrm { r } ^ { 3 } = \frac { \mathrm { T } ^ { 2 } \mathrm { GM } _ { \mathrm { E } } } { 4 \pi ^ { 2 } } \Rightarrow \mathrm { r } = \left( \frac { \mathrm { T } ^ { 2 } \mathrm { GM } _ { \mathrm { E } } } { 4 \pi ^ { 2 } } \right) ^ { 1 / 3 }$

Here, T = 24h = 24 × 60 × 60 s

$\mathrm { G } = 6.67 \times 10 ^ { - 11 } \mathrm { Nm } ^ { 2 } \mathrm {~kg} ^ { - 2 }$

$= 42.1 \times 10 ^ { 6 } \mathrm {~m}$