A satellite of mass m is at a distance a from a star of mass M. the speed of satellite is `u`. Suppose the low of universal gravity is `F=-G(Mm)/(r^(2.1))` instead of `F=-G(Mm)/(r^(2))` find the speed of the statellite when it is at a distance b from the star.

Options

A. `sqrt(u^(2)+2GM((1)/(b^(1.1))-(1)/(a^(1.1))))`
B. `sqrt(u^(2)+GM((1)/(a^(1.1))-(1)/(b^(1.1))))`
C. `sqrt(u^(2)+(2)/(1.1)GM((1)/(b^(1.1))-(1)/(a^(1.1))))`
D. `sqrt(u^(2)+(2)/(2.1)GM((1)/(b^(1.1))-(1)/(a^(1.1))))`

Explanation

To solve the problem, we need to find the speed of a satellite when it is at a distance \( b \) from a star of mass \( M \), given that the gravitational force follows the law \( F = -\frac{G M m}{r^{2.1}} \).

### Step-by-Step Solution:

1. **Identify the Work Done by the Gravitational Force**:
The work done \( W \) by the gravitational force when moving from distance \( a \) to \( b \) can be expressed as:
\[ W = \int_{a}^{b} F \, dr = \int_{a}^{b} -\frac{G M m}{r^{2.1}} \, dr \]

2. **Integrate the Force**:
To compute the integral, we rewrite it:
\[ W = -G M m \int_{a}^{b} r^{-2.1} \, dr \]
The integral of \( r^{-2.1} \) is:
\[ \int r^{-2.1} \, dr = \frac{r^{-1.1}}{-1.1} = -\frac{1}{1.1 r^{1.1}} + C \]
Therefore, we evaluate the definite integral:
\[ W = -G M m \left[ -\frac{1}{1.1 r^{1.1}} \right]_{a}^{b} = -G M m \left( -\frac{1}{1.1 b^{1.1}} + \frac{1}{1.1 a^{1.1}} \right) \]
Simplifying gives:
\[ W = \frac{G M m}{1.1} \left( \frac{1}{b^{1.1}} - \frac{1}{a^{1.1}} \right) \]

3. **Use the Work-Energy Theorem**:
According to the work-energy theorem, the work done is equal to the change in kinetic energy:
\[ W = K_f - K_i \]
Where \( K_i \) is the initial kinetic energy \( \frac{1}{2} m u^2 \) and \( K_f \) is the final kinetic energy \( \frac{1}{2} m v^2 \):
\[ \frac{G M m}{1.1} \left( \frac{1}{b^{1.1}} - \frac{1}{a^{1.1}} \right) = \frac{1}{2} m v^2 - \frac{1}{2} m u^2 \]

4. **Cancel Mass \( m \)**:
Since \( m \) appears in all terms, we can cancel it out:
\[ \frac{G M}{1.1} \left( \frac{1}{b^{1.1}} - \frac{1}{a^{1.1}} \right) = \frac{1}{2} v^2 - \frac{1}{2} u^2 \]

5. **Rearranging for \( v^2 \)**:
Rearranging gives:
\[ \frac{1}{2} v^2 = \frac{G M}{1.1} \left( \frac{1}{b^{1.1}} - \frac{1}{a^{1.1}} \right) + \frac{1}{2} u^2 \]
Multiplying through by 2:
\[ v^2 = \frac{2 G M}{1.1} \left( \frac{1}{b^{1.1}} - \frac{1}{a^{1.1}} \right) + u^2 \]

6. **Finding \( v \)**:
Finally, taking the square root gives:
\[ v = \sqrt{\frac{2 G M}{1.1} \left( \frac{1}{b^{1.1}} - \frac{1}{a^{1.1}} \right) + u^2} \]

### Final Answer:
\[ v = \sqrt{\frac{2 G M}{1.1} \left( \frac{1}{b^{1.1}} - \frac{1}{a^{1.1}} \right) + u^2} \]