Radius of orbit of satellite of earth is R. Its kinetic energy is proportional to:

Question

Radius of orbit of satellite of earth is R. Its kinetic energy is proportional to: (A)  (1/R)  (B)  (1/√R)  (C) R (D) (1/R3 / 2)

Tardigrade Admin · Accepted Answer

Kinetic energy of the satellite K E=(1/2) m v02 ....(1) Now putting the value of v0 is eq. (1), we get K E=(1/2) m(√(G M/R))2 =(1/2) (m G M/R) Hence, K E propto (1/R)

Q. Radius of orbit of satellite of earth is $R$ . Its kinetic energy is proportional to:

$\frac{1}{R}$

$\frac{1}{R}$

$R$

$\frac{1}{R ^{3/2}}$

Kinetic energy of the satellite
$K E = \frac{1}{2} m v_{0}^{2} .... (1)$
Now putting the value of $v_{0}$ is eq. (1), we get
$K E = \frac{1}{2} m (\frac{GM}{R})^{2}$
$= \frac{1}{2} \frac{m GM}{R}$
Hence, $K E \propto \frac{1}{R}$

Q. Radius of orbit of satellite of earth is R. Its kinetic energy is proportional to:

R1​

R​1​

R

R3/21​

Kinetic energy of the satellite KE=21​mv02​....(1) Now putting the value of v0​ is eq. (1), we get KE=21​m(RGM​​)2 =21​RmGM​ Hence, KE∝R1​

Q. Radius of orbit of satellite of earth is $R$ . Its kinetic energy is proportional to:

$\frac{1}{R}$

$\frac{1}{R}$

$R$

$\frac{1}{R ^{3/2}}$

Kinetic energy of the satellite
$K E = \frac{1}{2} m v_{0}^{2} .... (1)$
Now putting the value of $v_{0}$ is eq. (1), we get
$K E = \frac{1}{2} m (\frac{GM}{R})^{2}$
$= \frac{1}{2} \frac{m GM}{R}$
Hence, $K E \propto \frac{1}{R}$