Question

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

• A. $4\times {10}^4\text{km}$
• B. $3\times {10}^3\text{km}$
• C. $\frac{4}{3}\times {10}^3\text{km}$
• D. $1\times {10}^3\text{km}$

Answer 1

The correct option is C $\frac{4}{3}\times {10}^3\text{km}$

Given: Maximum velocity of satellite $=3\times {10}^4\text{m/s}$
minimum velocity of satellite $=1\times {10}^3\text{m/s}$
maximum distance of satellite from planet = $4\times {10}^4\text{km}$
minimum distance of satellite from planet = ?

As we know that the relation of maximum and minimum velocity with distance is given by
$\frac{r_{min}}{r_{max}}=\frac{v_{min}}{v_{max}}[\because v=wr]r_{min}=\frac{v_{min}}{v_{max}}\times r_{max}$
$=\frac{1\times {10}^3}{3\times {10}^4}\times 4\times {10}^{4}km=\frac{4}{3}\times {10}^3\text{km}$