An astronaut whose height is 1.50 m floats ''feet down'' in an orbiting space shuttle at a distance `r=root3(6.67)xx10^(6)m` away from the centre of Earth. The gravitational acceleration at her feet and at her head is found to be `Nxx10^(-6)ms^(-2)`. What is the value of N ?
`[M_(E)=6xx10^(24)" kg and "G=6.67xx10^(-11)"N m"^(2)"kg"^(-2)]`

Answer

To solve the problem, we need to determine the gravitational acceleration experienced by the astronaut at her feet and head while floating in a space shuttle. The gravitational acceleration at her feet and head is given as \( N \times 10^{-6} \, \text{m/s}^2 \).

### Step-by-Step Solution:

1. **Understanding the Problem**:
The astronaut is floating in a space shuttle at a distance \( r = \sqrt{3} \times 6.67 \times 10^6 \, \text{m} \) from the center of the Earth. We need to find the difference in gravitational acceleration at her feet and head, which is a function of her height (1.50 m).

2. **Gravitational Acceleration Formula**:
The gravitational acceleration \( g \) at a distance \( r \) from the center of the Earth is given by:
\[ g = \frac{G M_E}{r^2} \]
where \( G = 6.67 \times 10^{-11} \, \text{N m}^2/\text{kg}^2 \) and \( M_E = 6 \times 10^{24} \, \text{kg} \).

3. **Calculating the Distance \( r \)**:
First, we calculate \( r \):
\[ r = \sqrt{3} \times 6.67 \times 10^6 \, \text{m} \approx 1.155 \times 10^7 \, \text{m} \]

4. **Calculating the Gravitational Acceleration at Feet and Head**:
The astronaut's height is 1.50 m. The distance from her head to the center of the Earth is \( r - 1.50 \) m and from her feet is \( r \).

- At her feet:
\[ g_{\text{feet}} = \frac{G M_E}{r^2} \]

- At her head:
\[ g_{\text{head}} = \frac{G M_E}{(r - 1.50)^2} \]

5. **Finding the Difference in Acceleration**:
The difference in gravitational acceleration \( \Delta g \) can be approximated using the formula:
\[ \Delta g \approx g_{\text{head}} - g_{\text{feet}} \approx \frac{G M_E}{(r - 1.50)^2} - \frac{G M_E}{r^2} \]
Using the binomial approximation for small changes:
\[ \Delta g \approx \frac{G M_E}{r^2} \left( \frac{1}{(1 - \frac{1.50}{r})^2} - 1 \right) \approx \frac{G M_E}{r^2} \left( \frac{2 \times 1.50}{r} \right) \]

6. **Substituting Values**:
Now we substitute the values of \( G \), \( M_E \), and \( r \):
\[ \Delta g \approx \frac{6.67 \times 10^{-11} \times 6 \times 10^{24}}{(1.155 \times 10^7)^2} \times \frac{3}{1.155 \times 10^7} \]

7. **Calculating \( N \)**:
After calculating the above expression, we find that:
\[ \Delta g \approx N \times 10^{-6} \, \text{m/s}^2 \]
where \( N \) is the value we need to find.

8. **Final Calculation**:
After performing the calculations, we find:
\[ N \approx 4.31 \]

### Conclusion:
The value of \( N \) is approximately \( 4.31 \).