Question

an astronomical object has its mass 4 times the mass of earth and radius half of the radius of earth. If acceleration due to gravity at earth is g, find its value at the surface of the astronomical object.

Answer 1

Answer on Question #44393 – Physics – Mechanics | Kinematics | Dynamics

Question.

An astronomical object has its mass 4 times the mass of earth and radius half of the radius of earth. If acceleration due to gravity at earth is $g$, find its value at the surface of the astronomical object.

$$M = 4 M_E$$ $$R = 0.5 R_E$$ $$a_E = g$$ $$a = ?$$

Solution.

As we know from the gravitational law:

$$a = G \frac{M}{R^2}, \text{where}$$

$a$ is the acceleration at the surface of the astronomical object;

$G$ is the gravitational constant;

$M$ is the mass of the object;

$R$ is the radius of the object.

For Earth we have the following:

$$g = G \frac{M_E}{R_E^2}$$

Therefore,

$$a = G \frac{M}{R^2} = a = G \frac{4 M_E}{(0.5 R_E)^2} = 16 G \frac{M_E}{R_E^2} = 16 g$$

Answer.

$$a = 16 g$$

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