Question

An astronaut in a circular orbit around earth observes a celestial body moving in a lower circular orbit around earth in same plane as his orbit and in the same sense. He observes that the body moves at a speed of 5 m/s relative to himself when it is closest to him. The minimum distance between him and the body if he is moving at a speed of 5000 m/s is $\alpha$ km. Fill the value of $\frac{\alpha}{4}$ in OMR sheet. (Mass of earth = 6 × $10^{24}$ kg and round off the value of $\alpha$ to the nearest integer).

Answer 1

3

Answer 2

The problem describes two objects in circular orbits around Earth. The astronaut is in a higher orbit with speed $v_A = 5000$ m/s. The celestial body is in a lower orbit, meaning it must have a higher speed. The relative speed between them is $v_{rel} = 5$ m/s when they are closest. This relative speed is the difference between their orbital speeds, so the celestial body's speed is $v_B = v_A + v_{rel} = 5000 + 5 = 5005$ m/s.

The orbital speed $v$ for a circular orbit of radius $r$ around a central mass $M$ is given by $v = \sqrt{\frac{GM}{r}}$, which can be rearranged to find the radius: $r = \frac{GM}{v^2}$.

We are given the mass of Earth $M = 6 \times 10^{24}$ kg and the gravitational constant $G = 6.674 \times 10^{-11} \, \text{N m}^2/\text{kg}^2$. First, calculate the product $GM$: $GM = (6.674 \times 10^{-11} \, \text{N m}^2/\text{kg}^2) \times (6 \times 10^{24} \, \text{kg}) = 4.0044 \times 10^{14} \, \text{m}^3/\text{s}^2$.

Now, calculate the orbital radii for the astronaut ($r_A$) and the celestial body ($r_B$): $r_A = \frac{GM}{v_A^2} = \frac{4.0044 \times 10^{14} \, \text{m}^3/\text{s}^2}{(5000 \, \text{m/s})^2} = \frac{4.0044 \times 10^{14}}{25 \times 10^6} \, \text{m} = 1.60176 \times 10^7 \, \text{m}$. $r_B = \frac{GM}{v_B^2} = \frac{4.0044 \times 10^{14} \, \text{m}^3/\text{s}^2}{(5005 \, \text{m/s})^2} = \frac{4.0044 \times 10^{14}}{25050025} \, \text{m} \approx 1.6005570385 \times 10^7 \, \text{m}$.

The minimum distance between them occurs when they are aligned radially from Earth. This minimum distance is the difference between their orbital radii: $\Delta r = r_A - r_B = (1.60176 \times 10^7 \, \text{m}) - (1.6005570385 \times 10^7 \, \text{m})$ $\Delta r = (1.60176 - 1.6005570385) \times 10^7 \, \text{m} = 0.0012029615 \times 10^7 \, \text{m} = 12029.615 \, \text{m}$.

The problem states this minimum distance is $\alpha$ km. Convert meters to kilometers: $\alpha = \frac{12029.615 \, \text{m}}{1000 \, \text{m/km}} = 12.029615 \, \text{km}$.

Rounding $\alpha$ to the nearest integer gives $\alpha = 12$ km.

Finally, the value to be filled in the OMR sheet is $\frac{\alpha}{4}$: $\frac{\alpha}{4} = \frac{12}{4} = 3$.