Satellite Orbit Accelerations

Satellite Orbit Accelerations

Introduction

We are given two satellites, labeled 1 and 2, orbiting the Earth in circular orbits. The radius of satellite 2's orbit, $R<em>2$, is twice the radius of satellite 1's orbit, $R</em>1$. That is, $R<em>2 = 2R</em>1$. Our task is to determine how their accelerations, $a<em>2$ and $a</em>1$, are related.

Relevant Formula

The centripetal acceleration $a$ of an object in circular motion is given by the formula:

$a = \frac{v^2}{R}$

where $v$ is the orbital speed and $R$ is the radius of the circular orbit.

We can also express the orbital speed $v$ in terms of the gravitational constant $G$, the mass of the Earth $M$, and the radius of the orbit $R$:

$v = \sqrt{\frac{GM}{R}}$

Substituting this expression for $v$ into the centripetal acceleration formula, we get:

$a = \frac{(\sqrt{\frac{GM}{R}})^2}{R} = \frac{\frac{GM}{R}}{R} = \frac{GM}{R^2}$

Thus, the acceleration is inversely proportional to the square of the radius.

Applying the Formula to the Satellites

For satellite 1, the acceleration $a_1$ is:

$a<em>1 = \frac{GM}{R</em>1^2}$

For satellite 2, the acceleration $a_2$ is:

$a<em>2 = \frac{GM}{R</em>2^2}$

We are given that $R<em>2 = 2R</em>1$. Substituting this into the equation for $a_2$, we get:

$a<em>2 = \frac{GM}{(2R</em>1)^2} = \frac{GM}{4R_1^2}$

Determining the Relationship Between $a<em>1$ and $a</em>2$

Now we can express $a<em>2$ in terms of $a</em>1$:

$a<em>2 = \frac{GM}{4R</em>1^2} = \frac{1}{4} \cdot \frac{GM}{R<em>1^2} = \frac{1}{4} a</em>1$

Therefore, $a<em>2 = \frac{a</em>1}{4}$.

Conclusion

The acceleration of satellite 2 is one-fourth the acceleration of satellite 1. Thus $a<em>2 = a</em>1/4$. This corresponds to option E.