Question

A satellite is plad in a circular orbit around the earth at an altitude of 1000 km. The time period of the satellite in minutes is approximately (mass of the earth $= 6 \times 10^{24}$ kg, radius of the earth $= 6.4 \times 10^6$ m, $G = 6.67 \times 10^{-11}$ Nm$^2$kg$^{-2}$)

• A. $105$
• B. $200$
• C. $120$
• D. $62$

Hint:

Time Period of Satellite:

• Use $T = 2\pi \sqrt\fracr^3GM$, with $r = R_E + h$.
• Always convert km to meters before substitution.
• Final answer in seconds can be converted to minutes by dividing by 60.

Answer 1

The Correct Option is A

Total orbital radius $r = R_E + h = 6.4 \times 10^6 + 1 \times 10^6 = 7.4 \times 10^6$ m.
Kepler’s 3rd law: $T = 2\pi \sqrt{\frac{r^3}{GM}}$. Compute $r^3 = (7.4)^3 \times 10^{18} = 405.2 \times 10^{18}$.
$GM = 6.67 \times 10^{-11} \times 6 \times 10^{24} = 40.02 \times 10^{13}$.
Then, $\frac{r^3}{GM} \approx \frac{405.2 \times 10^{18}}{40.02 \times 10^{13}} = 10.13 \times 10^5$.
$T = 2\pi \sqrt{10.13 \times 10^5} \approx 2\pi \times 1006.25 = 6323$ s $= 105.4$ minutes. Hence, answer is 105 minutes.

Answer 2

Step 1: Understand the problem
We need to calculate the time period (T) of a satellite orbiting the Earth in a circular orbit at an altitude of 1000 km.

Step 2: Given data
- Mass of the Earth, M = \(6 \times 10^{24}\) kg
- Radius of the Earth, R = \(6.4 \times 10^{6}\) m
- Altitude of satellite, h = 1000 km = \(1 \times 10^{6}\) m
- Gravitational constant, G = \(6.67 \times 10^{-11}\) Nm²/kg²

Step 3: Calculate the orbital radius
The radius of the orbit, r = R + h = \(6.4 \times 10^{6} + 1 \times 10^{6} = 7.4 \times 10^{6}\) m

Step 4: Use the formula for orbital period
The orbital period is given by:
\[ T = 2\pi \sqrt{\frac{r^3}{GM}} \]

Step 5: Substitute values and calculate
\[ T = 2\pi \sqrt{\frac{(7.4 \times 10^{6})^3}{6.67 \times 10^{-11} \times 6 \times 10^{24}}} \]
Calculate inside the square root:
\[ r^3 = (7.4)^3 \times 10^{18} = 405.224 \times 10^{18} \]
\[ GM = 6.67 \times 10^{-11} \times 6 \times 10^{24} = 4.002 \times 10^{14} \]
So,
\[ \frac{r^3}{GM} = \frac{405.224 \times 10^{18}}{4.002 \times 10^{14}} = 1.012 \times 10^{6} \]
Taking square root:
\[ \sqrt{1.012 \times 10^{6}} \approx 1006 \]
Therefore,
\[ T = 2 \pi \times 1006 \approx 6321 \text{ seconds} \]
Convert to minutes:
\[ \frac{6321}{60} \approx 105 \text{ minutes} \]

Step 6: Conclusion
The time period of the satellite in its orbit is approximately 105 minutes.