Keplers Law

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New answer posted•a year ago

Keplers Law Class 11th

Two artificial satellites of masses $m_{1}$ and $m_{2}$ are moving with speeds $v_{1}$ and $v_{2}$ in orbits of radii $r_{1}$ and $r_{2}$ , respectively. If r₁>r₂ then which of the following statements is true

5 Views|Posted a year ago

Asked by Raj Pandey

Answered by

Raj Pandey | Contributor-Level 9

a year ago

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New answer posted•a year ago

Keplers Law Class 11th

Two planets A and B of equal mass are having their period of revolutions T_A and T_B such that T_A = 2T_B. These planets are revolving in the circular orbits or radii r_A and r_B respectively. Which out of the following would be the correct relationship of their orbits?

3 Views|Posted a year ago

Asked by Raj Pandey

Answered by

Raj Pandey | Contributor-Level 9

a year ago

According to kepler’s third law of time period –

${(\frac{T_{A}}{T_{B}})}^{2} = {(\frac{r_{A}}{r_{B}})}^{3} \Rightarrow \frac{r_{A}}{r_{B}} = 4^{1 / 3}$

$\Rightarrow r_{4}^{3} = 4 r_{B}^{3}$

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New answer posted•a year ago

Keplers Law Class 11th

The time period of satellite in a circular orbit of radius R is T. The period of another satellite in a circular orbit of radius 9R is:

6 Views|Posted a year ago

Asked by Raj Pandey

Answered by

Raj Pandey | Contributor-Level 9

a year ago

T² ∝ R³
⇒ (T? /T? )² = (R? /R? )³ ⇒ T? = 9³/² T = 27T

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New answer posted•a year ago

Keplers Law Class 11th

A satellite is launched into a circular orbit of radius R around earth, while a second satellite is launched into a circular orbit of radius 1.02R. The percentage difference in the time periods of the two satellites is:

7 Views|Posted a year ago

Asked by Vishal Baghel

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

Time period of Satellite = T = $\frac{2 π R}{v} = {(\frac{4 π^{2} R^{3}}{G M})}^{\frac{1}{2}}$

$\Rightarrow T α R^{\frac{3}{2}} \Rightarrow \frac{T_{2}}{T_{1}} = \frac{{(1.02 R)}^{\frac{3}{2}}}{R^{\frac{3}{2}}} = {(1.02)}^{\frac{3}{2}} = {(1 + 0.02)}^{\frac{3}{2}}$

The percentage difference in the time periods of the two satellites = $\frac{Δ T}{T_{1}}$ × 100

= (0.03) × 100 = 3%

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Keplers Law

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Two artificial satellites of masses m 1 and m 2 are moving with speeds v 1 and v 2 in orbits of radii r 1 and r 2 , respectively. If r1>r2 then which of the following statements is true

Two planets A and B of equal mass are having their period of revolutions TA and TB such that TA = 2TB. These planets are revolving in the circular orbits or radii rA and rB respectively. Which out of the following would be the correct relationship of their orbits?

The time period of satellite in a circular orbit of radius R is T. The period of another satellite in a circular orbit of radius 9R is:

A satellite is launched into a circular orbit of radius R around earth, while a second satellite is launched into a circular orbit of radius 1.02R. The percentage difference in the time periods of the two satellites is:

Two artificial satellites of masses $m_{1}$ and $m_{2}$ are moving with speeds $v_{1}$ and $v_{2}$ in orbits of radii $r_{1}$ and $r_{2}$ , respectively. If r₁>r₂ then which of the following statements is true

Two planets A and B of equal mass are having their period of revolutions T_A and T_B such that T_A = 2T_B. These planets are revolving in the circular orbits or radii r_A and r_B respectively. Which out of the following would be the correct relationship of their orbits?

The time period of satellite in a circular orbit of radius R is T. The period of another satellite in a circular orbit of radius 9R is: