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 calculate the gravitational acceleration at the feet and head of the astronaut and find the difference between the two. We will then express this difference in terms of \( N \times 10^{-6} \, \text{m/s}^2 \).

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Height of the astronaut, \( h = 1.50 \, \text{m} \)
- Distance from the center of the Earth, \( r = \sqrt{3} \times 6.67 \times 10^6 \, \text{m} \)
- Mass of the Earth, \( M_E = 6 \times 10^{24} \, \text{kg} \)
- Gravitational constant, \( G = 6.67 \times 10^{-11} \, \text{N m}^2/\text{kg}^2 \)

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

3. **Calculate the Gravitational Acceleration at the Feet:**
The gravitational acceleration \( g \) at a distance \( r \) from the center of the Earth is given by:
\[ g = \frac{G M_E}{r^2} \]
Substituting the values:
\[ g_{\text{feet}} = \frac{(6.67 \times 10^{-11}) \times (6 \times 10^{24})}{(1.155 \times 10^7)^2} \]

4. **Calculate \( r^2 \):**
\[ r^2 = (1.155 \times 10^7)^2 \approx 1.333 \times 10^{14} \, \text{m}^2 \]

5. **Substituting into the Equation for \( g_{\text{feet}} \):**
\[ g_{\text{feet}} = \frac{(6.67 \times 10^{-11}) \times (6 \times 10^{24})}{1.333 \times 10^{14}} \approx \frac{4.002 \times 10^{14}}{1.333 \times 10^{14}} \approx 3.003 \, \text{m/s}^2 \]

6. **Calculate the Gravitational Acceleration at the Head:**
The distance from the center of the Earth at the head of the astronaut is \( r + h \):
\[ g_{\text{head}} = \frac{G M_E}{(r + h)^2} \]
\[ g_{\text{head}} = \frac{(6.67 \times 10^{-11}) \times (6 \times 10^{24})}{(1.155 \times 10^7 + 1.5)^2} \]

7. **Calculate \( (r + h)^2 \):**
\[ r + h \approx 1.155 \times 10^7 + 1.5 \approx 1.155 \times 10^7 \, \text{m} \quad (\text{since } h \text{ is negligible compared to } r) \]
So, \( (r + h)^2 \approx (1.155 \times 10^7)^2 \approx 1.333 \times 10^{14} \, \text{m}^2 \) (approximately the same as \( r^2 \)).

8. **Calculate \( g_{\text{head}} \):**
\[ g_{\text{head}} \approx g_{\text{feet}} \quad (\text{since } h \text{ is negligible}) \]

9. **Calculate the Change in Gravitational Acceleration:**
The difference in gravitational acceleration between the feet and the head:
\[ \Delta g = g_{\text{head}} - g_{\text{feet}} \approx -\frac{2 G M_E h}{r^3} \]
Substituting the values:
\[ \Delta g \approx -\frac{2 \times (6.67 \times 10^{-11}) \times (6 \times 10^{24}) \times (1.5)}{(1.155 \times 10^7)^3} \]

10. **Calculate \( r^3 \):**
\[ r^3 \approx (1.155 \times 10^7)^3 \approx 1.54 \times 10^{21} \, \text{m}^3 \]

11. **Final Calculation of \( \Delta g \):**
\[ \Delta g \approx -\frac{2 \times (6.67 \times 10^{-11}) \times (6 \times 10^{24}) \times (1.5)}{1.54 \times 10^{21}} \approx -1.8 \times 10^{-4} \, \text{m/s}^2 \]

12. **Expressing in Terms of \( N \):**
\[ \Delta g \approx -1.8 \times 10^{-4} \, \text{m/s}^2 = -180 \times 10^{-6} \, \text{m/s}^2 \]
Thus, the value of \( N \) is \( 180 \).

### Final Answer:
\[ N = 180 \]