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
B
3×103km
C
4/3×103km
D
1×103km

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

Suggest Corrections

0

Similar questions
View More

People also searched for
View More