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Question

A satellite moves in elliptical orbit about a planet. Its maximum and minimum velocities of satellites are $$\displaystyle 3\times { 10 }^{ 4 }{ m }/{ s }$$ and $$\displaystyle 1\times { 10 }^{ 3 }{ m }/{ s }$$ respectively. What is the minimum distance of satellite from planet is maximum distance if $$\displaystyle 4\times { 10 }^{ 4 }km$$?


A
4×103km
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B
3×103km
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C
4/3×103km
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D
1×103km
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Solution

The correct option is D $$\displaystyle { 4 }/{ 3 }\times { 10 }^{ 3 }km$$
Maximum distance of satellite  $$r_{max} = 4\times 10^4 \ km$$
Velocity of satellite at this distance $$v_{min} = 1\times 10^3 \ m/s$$
Let the minimum distance of satellite be $$r_{min}$$. 
Velocity of satellite at this distance $$v_{max} = 3\times 10^3 \ m/s$$
Since angular momentum of satellite is conserved i.e.  $$mvr = $$ constant
$$\therefore$$  $$v_{max} r_{min} = v_{min} r_{max}$$
$$(3\times 10^3)\times r_{min} = (1\times 10^3)\times (4\times 10^4)$$
$$\implies$$  $$r_{min} = 4\times 10^3  \ km$$

Physics

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