Satellite Energy in Elliptic Orbit

The problem describes a satellite of mass \({\rm{m}}\) orbiting the Earth of mass \({\rm{M}}\) in an elliptic trajectory with a semi-major axis \({\rm{a}}\). We are given a specific condition at a radial distance \({\rm{r}} = {\rm{r_0}}\) from the Earth's center: the kinetic energy is equal to half the magnitude of the total energy. We are also given the formula for the total energy in such an orbit, \({\rm{E}} = - \frac{{{\rm{GMm}}}}{{{\rm{2a}}}}\). We need to find the ratio \({\rm{r_0}}/{\rm{a}}\).

Understanding Orbital Energy Components

The total energy \({\rm{E}}\) of a satellite in orbit is the sum of its kinetic energy \({\rm{K}}\) and its potential energy \({\rm{U}}\).

• Total Energy: \({\rm{E}} = {\rm{K}} + {\rm{U}}\)
• Potential Energy at a radial distance \({\rm{r}}\): \({\rm{U}}({\rm{r}}) = - \frac{{{\rm{GMm}}}}{{\rm{r}}}\)

From the total energy formula \({\rm{E}} = - \frac{{{\rm{GMm}}}}{{{\rm{2a}}}}\), we can find the kinetic energy \({\rm{K}}\) at any radial distance \({\rm{r}}\) using \({\rm{K}}({\rm{r}}) = {\rm{E}} - {\rm{U}}({\rm{r}})\).

\({\rm{K}}({\rm{r}}) = - \frac{{{\rm{GMm}}}}{{{\rm{2a}}}} - \left(- \frac{{{\rm{GMm}}}}{{\rm{r}}}\right) = \frac{{{\rm{GMm}}}}{{\rm{r}}} - \frac{{{\rm{GMm}}}}{{{\rm{2a}}}}\)

Applying the Kinetic Energy Condition

We are given that at \({\rm{r}} = {\rm{r_0}}\), the kinetic energy is equal to half the magnitude of the total energy. Since \({\rm{E}}\) is negative for a bound orbit (\({\rm{E}} = - \frac{{{\rm{GMm}}}}{{{\rm{2a}}}}\)), its magnitude is \(|{\rm{E}}| = -{\rm{E}}\).

The condition is: \({\rm{K}}({\rm{r_0}}) = \frac{1}{2} |{\rm{E}}|\)

Substituting the expressions for \({\rm{K}}({\rm{r_0}})\) and \(|{\rm{E}}|\):

\( \frac{{{\rm{GMm}}}}{{{\rm{r_0}}}} - \frac{{{\rm{GMm}}}}{{{\rm{2a}}}} = \frac{1}{2} \left|- \frac{{{\rm{GMm}}}}{{{\rm{2a}}}}\right| \)

\( \frac{{{\rm{GMm}}}}{{{\rm{r_0}}}} - \frac{{{\rm{GMm}}}}{{{\rm{2a}}}} = \frac{1}{2} \left( \frac{{{\rm{GMm}}}}{{{\rm{2a}}}} \right) \)

\( \frac{{{\rm{GMm}}}}{{{\rm{r_0}}}} - \frac{{{\rm{GMm}}}}{{{\rm{2a}}}} = \frac{{{\rm{GMm}}}}{{{\rm{4a}}}} \)

Solving for the Ratio \({\rm{r_0}}/{\rm{a}}\)

We can divide the entire equation by \({\rm{GMm}}\) (assuming \({\rm{G}}\), \({\rm{M}}\), \({\rm{m}}\) are not zero):

\( \frac{1}{{{\rm{r_0}}}} - \frac{1}{{{\rm{2a}}}} = \frac{1}{{{\rm{4a}}}} \)

Now, let's isolate the term with \({\rm{r_0}}\):

\( \frac{1}{{{\rm{r_0}}}} = \frac{1}{{{\rm{4a}}}} + \frac{1}{{{\rm{2a}}}} \)

To add the fractions on the right side, we find a common denominator, which is \({\rm{4a}}\):

\( \frac{1}{{{\rm{r_0}}}} = \frac{1}{{4a}} + \frac{2}{{4a}} \)

\( \frac{1}{{{\rm{r_0}}}} = \frac{1+2}{4a} = \frac{3}{{{\rm{4a}}}} \)

Now, we can solve for \({\rm{r_0}}\) by taking the reciprocal of both sides:

\( {\rm{r_0}} = \frac{{{\rm{4a}}}}{3} \)

Finally, we find the required ratio \({\rm{r_0}}/{\rm{a}}\):

\( \frac{{{\rm{r_0}}}}{{\rm{a}}} = \frac{{{\rm{4a}}/3}}{{\rm{a}}} = \frac{4}{3} \)

The numerical value of the ratio is \(4/3 \approx 1.333\).

Matching with Options

The calculated value \(1.333...\) is nearest to option 1, which is 1.33.