To measure the mass of a planet with the same radius as Earth, an astronaut drops an object from rest (relative to the planet) from an altitude of one radius above the surface. When the object hits its speed is 4 times what it would be if the same experiment were carried out for Earth. In units of Earth masses, the mass of the planet is:

Question

Quizplus · Accepted Answer

Using conservation of energy, we know that the initial gravitational potential energy of the object is equal to its final kinetic energy when it hits the surface.

For Earth, the initial potential energy is:
$$U_i = \frac{Gm_eM_e}{2R_e}$$
where $m_e$ is the mass of the object, $M_e$ is the mass of Earth, and $R_e$ is the radius of Earth.

The final kinetic energy is:
$$K_f = \frac{1}{2}m_ev_e^2$$
where $v_e$ is the final velocity of the object when it hits Earth's surface.

Setting $U_i = K_f$, we can solve for $v_e$:
$$v_e = \sqrt{\frac{2G M_e}{R_e}}$$

For the planet with the same radius as Earth, we can set up the same equations with $M_p$ as the mass of the planet:
$$U_i = \frac{Gm_pM_p}{2R_e}$$
and
$$K_f = \frac{1}{2}m_ev_p^2$$
where $v_p$ is the final velocity of the object when it hits the planet's surface.

Setting $U_i = K_f$ again, and using the fact that $v_p = 4v_e$, we get:
$$\frac{Gm_pM_p}{2R_e} = \frac{1}{2}m_e(4v_e)^2$$
Simplifying and solving for $M_p$, we get:
$$M_p = 16\frac{m_e}{M_e}$$
Since we are measuring the mass of the planet in units of Earth masses, we can substitute $m_e$ for the mass of the object in Earth masses (which is 1):
$$M_p = 16 M_\oplus$$
where $M_\oplus$ is the mass of Earth. Therefore, the answer is D) 16.

To Measure the Mass of a Planet with the Same