The orbital velocity of an artificial satellite in a circular orbit very close to Earth is v. The velocity of a geosynchronous satellite orbiting in a circular orbit at an altitude of 6 R from Earth's surface will be

Question

The orbital velocity of an artificial satellite in a circular orbit very close to Earth is v. The velocity of a geosynchronous satellite orbiting in a circular orbit at an altitude of 6 R from Earth's surface will be (A) (v/√7) (B) (v/√6) (C) v (D) √6 v

Tardigrade Admin · Accepted Answer

For a satellite, (m v2/R)=(G M m/R2) ; v2=(G M/R) ((v2/v1))2=((R1/R2)) ⇒ ((v prime/v))2=((R/R+6 R)) ⇒ v prime=(v/√7)

Q. The orbital velocity of an artificial satellite in a circular orbit very close to Earth is $v$ . The velocity of a geosynchronous satellite orbiting in a circular orbit at an altitude of $6 R$ from Earth's surface will be

$\frac{v}{7}$

$\frac{v}{6}$

$v$

$6 v$

For a satellite, $\frac{m v ^{2}}{R} = \frac{GM m}{R ^{2}}; v^{2} = \frac{GM}{R}$
$(\frac{v _{2}}{v _{1}})^{2} = (\frac{R _{1}}{R _{2}}) \Rightarrow (\frac{v ^{'}}{v})^{2} = (\frac{R}{R + 6 R})$
$\Rightarrow v^{'} = \frac{v}{7}$

Q. The orbital velocity of an artificial satellite in a circular orbit very close to Earth is v. The velocity of a geosynchronous satellite orbiting in a circular orbit at an altitude of 6R from Earth's surface will be

7​v​

6​v​

v

6​v

For a satellite, Rmv2​=R2GMm​;v2=RGM​ (v1​v2​​)2=(R2​R1​​)⇒(vv′​)2=(R+6RR​) ⇒v′=7​v​

Q. The orbital velocity of an artificial satellite in a circular orbit very close to Earth is $v$ . The velocity of a geosynchronous satellite orbiting in a circular orbit at an altitude of $6 R$ from Earth's surface will be

$\frac{v}{7}$

$\frac{v}{6}$

$v$

$6 v$

For a satellite, $\frac{m v ^{2}}{R} = \frac{GM m}{R ^{2}}; v^{2} = \frac{GM}{R}$
$(\frac{v _{2}}{v _{1}})^{2} = (\frac{R _{1}}{R _{2}}) \Rightarrow (\frac{v ^{'}}{v})^{2} = (\frac{R}{R + 6 R})$
$\Rightarrow v^{'} = \frac{v}{7}$