A spaceship of mass 70 kg is revolving in a circular orbit at a height of 1000 km from the surface of earth. What will be acceleration due to gravity at any point along the path of spaceship ?

Answer

To find the acceleration due to gravity at a height of 1000 km above the Earth's surface, we can use the formula for gravitational acceleration:

\[ g' = \frac{GM}{r^2} \]

Where:
- \( g' \) is the acceleration due to gravity at height,
- \( G \) is the universal gravitational constant, approximately \( 6.67 \times 10^{-11} \, \text{N m}^2/\text{kg}^2 \),
- \( M \) is the mass of the Earth, approximately \( 6 \times 10^{24} \, \text{kg} \),
- \( r \) is the distance from the center of the Earth to the object.

### Step 1: Calculate the radius \( r \)
The radius \( r \) is the sum of the Earth's radius and the height of the spaceship above the Earth's surface.

- The average radius of the Earth \( R \) is approximately \( 6400 \, \text{km} \).
- The height of the spaceship \( h \) is \( 1000 \, \text{km} \).

So,

\[ r = R + h = 6400 \, \text{km} + 1000 \, \text{km} = 7400 \, \text{km} \]

Convert this to meters:

\[ r = 7400 \times 10^3 \, \text{m} = 7.4 \times 10^6 \, \text{m} \]

### Step 2: Substitute values into the formula
Now, we can substitute the values into the gravitational acceleration formula:

\[ g' = \frac{(6.67 \times 10^{-11}) \times (6 \times 10^{24})}{(7.4 \times 10^6)^2} \]

### Step 3: Calculate \( r^2 \)
Calculate \( r^2 \):

\[ r^2 = (7.4 \times 10^6)^2 = 54.76 \times 10^{12} \, \text{m}^2 \]

### Step 4: Substitute \( r^2 \) back into the formula
Now substitute \( r^2 \) back into the equation for \( g' \):

\[ g' = \frac{(6.67 \times 10^{-11}) \times (6 \times 10^{24})}{54.76 \times 10^{12}} \]

### Step 5: Calculate \( g' \)
Now, calculate the value:

1. Calculate the numerator:

\[ 6.67 \times 10^{-11} \times 6 \times 10^{24} = 40.02 \times 10^{13} \]

2. Now divide by the denominator:

\[ g' = \frac{40.02 \times 10^{13}}{54.76 \times 10^{12}} \approx 7.29 \, \text{m/s}^2 \]

### Conclusion
Thus, the acceleration due to gravity at the height of 1000 km above the Earth's surface is approximately:

\[ g' \approx 7.29 \, \text{m/s}^2 \]