A satellite is moving around the earth with speed v in a circular orbit of radius r. If the orbit radius is decreased by 1%, its speed will

The correct option is: Increase by 0.5%

Step-by-Step Derivation and Explanation:

1. Recall the formula for orbital speed:
A satellite of mass $m$ orbiting the Earth of mass $M$ in a circular orbit of radius $r$ experiences a gravitational force that provides the necessary centripetal force for its circular motion.
The equation is:
$\frac{m{v}^2}{r}=\frac{GMm}{r^2}$
Solving for the orbital speed $v$, we get:
$v=\sqrt{\frac{GM}{r}}$
where $G$ is the universal gravitational constant and $M$ is the mass of the Earth.

2. Express the relationship using proportionality:
Since $G$ and $M$ are constants, the orbital speed $v$ is inversely proportional to the square root of the orbit radius $r$:
$v\propto r^{-1/2}$

3. Use the approximation for small percentage changes:
Taking the natural logarithm on both sides of $v={\left(GM\right)}^{1/2}{r}^{-1/2}$, we get:
$\ln \left(v\right)=\frac{1}{2}\ln \left(GM\right)-\frac{1}{2}\ln \left(r\right)$
Differentiating both sides to find the fractional change, we obtain:
$\frac{dv}{v}=-\frac{1}{2}\frac{dr}{r}$
Multiplying by 100 to express this in percentage terms:
$(\frac{dv}{v}\times 100)=-\frac{1}{2}(\frac{dr}{r}\times 100)$

4. Substitute the given values:
We are given that the orbit radius is decreased by 1%, which means:
$\frac{dr}{r}\times 100=-1\%$
Substituting this value into our percentage change equation:
$(\frac{dv}{v}\times 100)=-\frac{1}{2}(-1\%)=+0.5\%$

5. Conclusion:
A positive sign indicates an increase. Therefore, if the orbit radius decreases by 1%, the speed of the satellite will increase by 0.5%.