Two satellites A and B revolve around a planet in two coplanar circular orbits in the same sense with radii 104km and 2 -104 km respectively- Time period of A is 28 hours- What is time period of anothersatellite-

From Kepler's first law, we know that T ^ 2 α R ^ 3 Thus, Tα R ^ 3/2 T _ A T _ B = R _ A ^ 3/2 R _ B ^ 3/2 #160; T _ B = 2 ^ 3/2 T _ A ...

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Physics

Satellites

Two satellite...

Question

Two satellites $A$ and $B$ revolve around a planet in two coplanar circular orbits in the same sense with radii $10^{4}$ km and $2 \times 104$ km respectively. Time period of $A$ is $28$ hours. What is time period of another
satellite?

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Solution

From Kepler's first law, we know that

$T^{2} α R^{3}$

Thus, $T α R^{3 / 2}$

$\frac{T_{A}}{T_{B}} = \frac{R_{A}^{3 / 2}}{R_{B}^{3 / 2}}$

$T_{B} = 2^{3 / 2} T_{A} = 79.2 h o u r s$

Answer is $79.2 h o u r s$

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Two satellites A and B revolve around a planet in two coplanar circular orbits in the same sense with radii 104km and 2×104 km respectively. Time period of A is 28 hours. What is time period of anothersatellite?

From Kepler's first law, we know thatT2αR3Thus, TαR3/2TATB=R3/2AR3/2BTB=23/2TA=79.2hoursAnswer is 79.2hours

Two satellites $A$ and $B$ revolve around a planet in two coplanar circular orbits in the same sense with radii $10^{4}$ km and $2 \times 104$ km respectively. Time period of $A$ is $28$ hours. What is time period of another
satellite?

From Kepler's first law, we know that

$T^{2} α R^{3}$

Thus, $T α R^{3 / 2}$

$\frac{T_{A}}{T_{B}} = \frac{R_{A}^{3 / 2}}{R_{B}^{3 / 2}}$

$T_{B} = 2^{3 / 2} T_{A} = 79.2 h o u r s$

Answer is $79.2 h o u r s$