India's Mangalyan was sent to the Mars by launching it into a transfer orbit EOM around then sun. It leaves the earth at E and meets Mars at M. If the semi-major axis of Earth's orbit is `a_(e) = 1.5 xx 10^(11)` m, that of Mars orbit `a_(m) = 2.28 xx 10^(11)` m , take Kepler's laws give the estimate of time for Mangalyan to reach Mars from Earth to be close to :
` (##DSH_NTA_JEE_MN_PHY_C07_E01_008_Q01.png" width="80%">

Options

A. 500 days
B. 320 days
C. 260 days
D. 220 days

Answer

To estimate the time for Mangalyaan to reach Mars from Earth using Kepler's laws, we can follow these steps:

### Step 1: Determine the semi-major axis of the transfer orbit
The semi-major axis \( A \) of the elliptical transfer orbit can be calculated as the average of the semi-major axes of Earth's orbit \( a_e \) and Mars' orbit \( a_m \).

\[ A = \frac{a_e + a_m}{2} \]

Given:
- \( a_e = 1.5 \times 10^{11} \, \text{m} \)
- \( a_m = 2.28 \times 10^{11} \, \text{m} \)

Calculating \( A \):

\[ A = \frac{1.5 \times 10^{11} + 2.28 \times 10^{11}}{2} = \frac{3.78 \times 10^{11}}{2} = 1.89 \times 10^{11} \, \text{m} \]

### Step 2: Use Kepler's Third Law
Kepler's Third Law states that the square of the period \( T \) of an orbit is proportional to the cube of the semi-major axis \( R \) of the orbit:

\[ \frac{T^2}{T_e^2} = \frac{R^3}{a_e^3} \]

Where:
- \( T_e = 365 \, \text{days} \) (the period of Earth's orbit)
- \( R = 1.89 \times 10^{11} \, \text{m} \) (the semi-major axis of the transfer orbit)
- \( a_e = 1.5 \times 10^{11} \, \text{m} \)

### Step 3: Calculate \( T^2 \)
Substituting the values into Kepler's law:

\[ T^2 = T_e^2 \cdot \frac{R^3}{a_e^3} \]

Calculating \( a_e^3 \):

\[ a_e^3 = (1.5 \times 10^{11})^3 = 3.375 \times 10^{33} \, \text{m}^3 \]

Calculating \( R^3 \):

\[ R^3 = (1.89 \times 10^{11})^3 = 6.749 \times 10^{33} \, \text{m}^3 \]

Now substituting these into the equation:

\[ T^2 = 365^2 \cdot \frac{6.749 \times 10^{33}}{3.375 \times 10^{33}} \]

Calculating the fraction:

\[ \frac{6.749}{3.375} \approx 2 \]

Thus:

\[ T^2 \approx 365^2 \cdot 2 \]

Calculating \( T^2 \):

\[ T^2 \approx 365^2 \cdot 2 = 133225 \cdot 2 = 266450 \]

### Step 4: Calculate \( T \)
Taking the square root to find \( T \):

\[ T \approx \sqrt{266450} \approx 516 \, \text{days} \]

### Step 5: Adjust for the elliptical path
Since the transfer orbit is not a perfect circle, we can estimate that the time taken will be about half of the calculated period for a circular orbit. Thus, we divide by 2:

\[ T \approx \frac{516}{2} \approx 258 \, \text{days} \]

### Conclusion
The estimated time for Mangalyaan to reach Mars from Earth is approximately **260 days**.