Question

A body of mass 1 kg moves in an elliptical orbit with semi-major axis 1000 m and semi-minor axis 100 m. The orbital angular momentum is 100 kg m$^2$ s$^{-1}$. The time period of motion is ............. hours. (Specify answer up to two digits after the decimal point.)
 

Hint:

For elliptical orbits, period is found using area swept per unit time: $L = 2mA/T$.

Answer 1

Correct Answer: 1.7

Step 1: Use area law.
Kepler's second law: angular momentum $L = 2mA/T$, where $A$ is area of ellipse.

Step 2: Compute area of ellipse.
$A = \pi ab = \pi (1000)(100) = 100000\pi$.

Step 3: Substitute in the relation.
$L = 2mA/T $\Rightarrow$ T = \frac{2mA}{L}$.
Since $m = 1$ kg:
$T = \frac{2(100000\pi)}{100} = 2000\pi\ \text{seconds}$.

Step 4: Convert seconds to hours.
$2000\pi \approx 6283\ \text{s}$.
$\text{Hours} = 6283/3600 = 1.75\ \text{hours}$.