An artificial satellite is moving in a circular orbit around the earth, with a speed equal to half the magnitude of escape velocity from the Earth. The height of satellite above the surface of earth is (where R is the radius of the Earth)

Question

An artificial satellite is moving in a circular orbit around the earth, with a speed equal to half the magnitude of escape velocity from the Earth. The height of satellite above the surface of earth is (where R is the radius of the Earth) (A) R (B) 2R (C) 3R (D) 4R

Tardigrade Admin · Accepted Answer

(1/2)√(2 G M/R)=√(G M/R + h) (1/4)× (2 G M/R)= (G M/R + h) 2R = R + h h = R

Q. An artificial satellite is moving in a circular orbit around the earth, with a speed equal to half the magnitude of escape velocity from the Earth. The height of satellite above the surface of earth is (where $R$ is the radius of the Earth)

$R$

$2 R$

$3 R$

$4 R$

$\frac{1}{2} \frac{2 GM}{R} = \frac{GM}{R + h}$
$\frac{1}{4} \times \frac{2 GM}{R} = \frac{GM}{R + h}$
$2 R = R + h$
$h = R$

Q. An artificial satellite is moving in a circular orbit around the earth, with a speed equal to half the magnitude of escape velocity from the Earth. The height of satellite above the surface of earth is (where R is the radius of the Earth)

R

2R

3R

4R

21​R2GM​​=R+hGM​​ 41​×R2GM​=R+hGM​ 2R=R+h h=R

Q. An artificial satellite is moving in a circular orbit around the earth, with a speed equal to half the magnitude of escape velocity from the Earth. The height of satellite above the surface of earth is (where $R$ is the radius of the Earth)

$R$

$2 R$

$3 R$

$4 R$

$\frac{1}{2} \frac{2 GM}{R} = \frac{GM}{R + h}$
$\frac{1}{4} \times \frac{2 GM}{R} = \frac{GM}{R + h}$
$2 R = R + h$
$h = R$