Question

the magnitude of the gravitational field at earth’s surface is $g_e$. planet x has two times the radius and four times the mass of the earth. what is the magnitude of the gravitational field at the surface of planet x in terms of $g_e$? choose 1 answer: a $\frac{1}{4}g_e$ b $g_e$ c $2g_e$

Explanation:

Step1: Recall Gravitational Field Formula

The formula for the gravitational field \( g \) at the surface of a planet is \( g = \frac{GM}{R^2} \), where \( G \) is the gravitational constant, \( M \) is the mass of the planet, and \( R \) is the radius of the planet. For Earth, \( g_E = \frac{G M_E}{R_E^2} \).

Step2: Define Parameters for Planet X

Let the mass of Planet X be \( M_X = 4 M_E \) (four times Earth's mass) and the radius of Planet X be \( R_X = 2 R_E \) (two times Earth's radius).

Step3: Calculate \( g_X \)

Substitute \( M_X \) and \( R_X \) into the gravitational field formula for Planet X:
\( g_X = \frac{G M_X}{R_X^2} = \frac{G (4 M_E)}{(2 R_E)^2} \).

Simplify the denominator: \( (2 R_E)^2 = 4 R_E^2 \).
So, \( g_X = \frac{4 G M_E}{4 R_E^2} \).

The 4 in the numerator and denominator cancels out, leaving \( g_X = \frac{G M_E}{R_E^2} \), which is equal to \( g_E \).

Answer:

B. \( g_E \)