For a satellite moving in an orbit around the earth, the ratio of kinetic energy to potential energy is:

Question

For a satellite moving in an orbit around the earth, the ratio of kinetic energy to potential energy is: (A) 2 (B)  (1/2)  (C)  (1/√2)  (D)  √2

Tardigrade Admin · Accepted Answer

(G M m/R2)+m ω2 R=0 ∴ (G M m/R2)=m ω2 R K . E .=(1/2) I ω2=(1/2) m R2 ω2=(G M m/2 R) P . E .=-(G M m/R) ∴ K . E .=(|P . E .|/2) ∴ (K . E ./|P . E .|)=(1/2)

Q. For a satellite moving in an orbit around the earth, the ratio of kinetic energy to potential energy is:

$2$

$\frac{1}{2}$

$\frac{1}{2}$

$2$

$\frac{GM m}{R ^{2}} + m ω^{2} R = 0$
$∴ \frac{GM m}{R ^{2}} = m ω^{2} R$
$K . E . = \frac{1}{2} I ω^{2} = \frac{1}{2} m R^{2} ω^{2} = \frac{GM m}{2 R}$
$P . E . = - \frac{GM m}{R}$
$∴ K . E . = \frac{∣ P . E .∣}{2}$
$∴ \frac{K . E .}{∣ P . E .∣} = \frac{1}{2}$

Q. For a satellite moving in an orbit around the earth, the ratio of kinetic energy to potential energy is:

2

21​

2​1​

2​

R2GMm​+mω2R=0 ∴R2GMm​=mω2R K.E.=21​Iω2=21​mR2ω2=2RGMm​ P.E.=−RGMm​ ∴K.E.=2∣P.E.∣​ ∴∣P.E.∣K.E.​=21​

$2$

$\frac{1}{2}$

$\frac{1}{2}$

$2$

$\frac{GM m}{R ^{2}} + m ω^{2} R = 0$
$∴ \frac{GM m}{R ^{2}} = m ω^{2} R$
$K . E . = \frac{1}{2} I ω^{2} = \frac{1}{2} m R^{2} ω^{2} = \frac{GM m}{2 R}$
$P . E . = - \frac{GM m}{R}$
$∴ K . E . = \frac{∣ P . E .∣}{2}$
$∴ \frac{K . E .}{∣ P . E .∣} = \frac{1}{2}$