A satellite is in an elliptical orbit around Earth. At one point on the ellipse, it is 25 × 10⁶ m from the center of Earth and has a speed of 12,000 m/s. What is its speed when it is at another point on the ellipse, 95 × 10⁶ m from the center of Earth? The mass of Earth is 5.97 × 10²⁴ kg and G = 6.67 x 10^-11 N·m²/kg². A) 12000 m/s B) 11000 m/s C) 10000 m/s D) 9000 m/s E) 8000 m/s

Question

Quizplus · Accepted Answer

We can use conservation of energy to solve this problem. The total mechanical energy of the satellite is constant throughout its orbit and is given by:

$E = \frac{1}{2}mv^2 - \frac{GMm}{r}$

where $m$ is the mass of the satellite, $v$ is the speed of the satellite, $r$ is the distance between the satellite and the center of Earth, $G$ is the gravitational constant, and $M$ is the mass of Earth.

At the first point on the ellipse, we have:

$E_1 = \frac{1}{2}m(12000 	ext{ m/s})^2 - \frac{(6.67 	imes 10^{-11} 	ext{ N}\cdot	ext{m}^2/	ext{kg}^2)(5.97	imes10^{24}	ext{ kg})m}{25	imes10^6	ext{ m}}$

Simplifying this expression, we get:

$E_1 = -7.47	imes10^{10}m$

where we have used the fact that the potential energy is negative.

At the second point on the ellipse, we want to find the speed $v_2$ that gives the same total energy:

$E_2 = \frac{1}{2}m v_2^2 - \frac{(6.67 	imes 10^{-11} 	ext{ N}\cdot	ext{m}^2/	ext{kg}^2)(5.97	imes10^{24}	ext{ kg})m}{95	imes10^6	ext{ m}}$

Setting $E_1 = E_2$ and solving for $v_2$, we get:

$v_2 = \sqrt{\frac{-7.47	imes10^{10}m + (6.67 	imes 10^{-11} 	ext{ N}\cdot	ext{m}^2/	ext{kg}^2)(5.97	imes10^{24}	ext{ kg})m}{\frac{1}{2}m}}$

We can simplify this expression by canceling the mass $m$ and evaluating the terms inside the square root:

$v_2 = \sqrt{\frac{-7.47	imes10^{10} + (6.67 	imes 10^{-11} 	ext{ N}\cdot	ext{m}^2/	ext{kg}^2)(5.97	imes10^{24})}{\frac{1}{2}}} \approx \boxed{	extbf{(B) }11000	ext{ m/s}}$

Note that we did not need to know the eccentricity of the elliptical orbit to solve this problem.

A Satellite Is in an Elliptical Orbit Around Earth