The acceleration due to gravity near a planet's surface is known to be 3.20 .

Part A

If the escape speed from the planet is 9.54 , determine its radius.
Part B

Find the mass of the planet.
Part C

If a probe is launched from its surface with a speed twice the escape speed and then coasts outward, neglecting other nearby astronomical bodies, what will be its speed when it is very far from the planet? (Neglect any atmospheric effects also.)
Part D

Under these launch conditions, at what distance will its speed be equal to the escape speed?

To solve this problem, we will use the gravitational force equation and the concept of escape speed.

The gravitational force between two objects is given by the equation F = (G * m1 * m2) / r^2, where F is the gravitational force, G is the gravitational constant (approximately 6.67 x 10^-11 Nm^2/kg^2), m1 and m2 are the masses of the objects, and r is the distance between the centers of the objects.

For Part A: To determine the radius of the planet, we can use the formula for escape speed. The escape speed is given by the equation v_escape = √(2 * g * r), where v_escape is the escape speed, g is the acceleration due to gravity, and r is the radius of the planet. We can solve this equation for r by rearranging the terms:

r = (v_escape^2) / (2 * g)

Substituting the given values, we have:

r = (9.54^2) / (2 * 3.20)

Simplifying the equation gives:

r = 28.28 / 6.40

r ≈ 4.42 m/s

Therefore, the radius of the planet is approximately 4.42 m/s.

For Part B: To find the mass of the planet, we can rearrange the gravitational force equation:

F = (G * m1 * m2) / r^2

Solving for m2 (mass of the planet), we have:

m2 = (F * r^2) / (G * m1)

Since we are given the acceleration due to gravity (g) near the planet's surface, we can use the equation F = m1 * g, where m1 is the mass of an object near the planet's surface. Substituting this into the equation, we have:

m2 = (m1 * g * r^2) / (G * m1)

Simplifying the equation gives:

m2 = (g * r^2) / G

Substituting the given values, we have:

m2 = (3.20 * (4.42^2)) / (6.67 x 10^-11)

Simplifying the equation gives:

m2 ≈ 7.86 x 10^23 kg

Therefore, the mass of the planet is approximately 7.86 x 10^23 kg.

For Part C: Since the probe is launched with a speed twice the escape speed, its initial speed is 2 * 9.54 = 19.08 m/s. As the probe moves away from the planet, the gravitational influence decreases, and its speed will reduce as it coasts outward. When the probe is very far from the planet, the gravitational force becomes negligible, and the probe will have a constant speed.

Therefore, when the probe is very far from the planet, its speed will be approximately 19.08 m/s.

For Part D: To find the distance at which the probe's speed is equal to the escape speed, we can use the equation for escape speed. Rearranging the equation, we have:

r = (v_escape^2) / (2 * g)

Substituting the given values, we have:

r = (9.54^2) / (2 * 3.20)

Simplifying the equation gives:

r = 28.28 / 6.40

r ≈ 4.42 m/s

Therefore, the distance at which the probe's speed is equal to the escape speed is approximately 4.42 m/s from the planet's surface.