A planet is made of two distinct materials. From the core to R/2, the density of the material is 4000 kg/m3, and from R/2 to R the density is 3000 kg/m3. What is the gravity at the surface of the planet if R = 5000 km? A) 0.699 m/s² B) 3.66 m/s² C) 4.19 m/s² D) 4.38 m/s² E) 4.99 m/s²

Question

Quizplus · Accepted Answer

Using the formula for gravitational acceleration at the surface of a planet,

$$g = \frac{GM}{R^2}$$

where $G$ is the gravitational constant, $M$ is the mass of the planet, and $R$ is the radius of the planet.

To find the mass of the planet, we divide it into two parts, one with radius $R/2$ and density $4000$ kg/m$^3$, and the other with radius $R/2$ to $R$ and density $3000$ kg/m$^3$. We then use the formulas for the volume and mass of a sphere to find the mass of each part and add them together.

For the inner part of the planet:

$$V = \frac{4}{3}\pi\left(\frac{R}{2}ight)^3 = \frac{\pi}{6}R^3$$

$$M = ho V = 4000 \cdot \frac{\pi}{6}R^3 = \frac{2}{3}\pi R^3$$

For the outer part of the planet:

$$V = \frac{4}{3}\pi\left(R^3 - \left(\frac{R}{2}ight)^3ight) = \frac{7}{12}\pi R^3$$

$$M = ho V = 3000 \cdot \frac{7}{12}\pi R^3 = \frac{7}{4}\pi R^3$$

Adding the masses together:

$$M_{	ext{total}} = \frac{2}{3}\pi R^3 + \frac{7}{4}\pi R^3 = \frac{29}{12}\pi R^3 $$

Plugging in the values into the formula for gravitational acceleration, we get:

$$g = \frac{GM}{R^2} = \frac{G \cdot \frac{29}{12}\pi R^3}{R^2} = \frac{29}{12}G\pi R $$

We can now use the approximation that $G\approx 6.67 	imes 10^{-11}$ Nm$^2$/kg$^2$ to get:

$$g \approx 4.38 \frac{	ext{m}}{	ext{s}^2}$$

Therefore, the answer is D, 4.38 m/s$^2$.

A Planet Is Made of Two Distinct Materials