Question

An astronaut, whose mission is to go where no one has gone before, lands on a spherical planet in a distant galaxy.
As she stands on the surface of the planet, she releases a small rock from rest and finds that it takes the rock 0.420 s to fall 1.90 m.
A) If the radius of the planet is 7.40×107 m,
B) what is the mass of the planet?

Answer 1

Answer:

$M = 1.77 \times 10^{27} kg$

Explanation:

As we know that the that the Astronaut dropped a stone on the surface of the planet takes 0.420 s to fall a distance of d = 1.90 m

so we will have

$y = \frac{1}{2}g t^2$

$1.90 = \frac{1}{2}g(0.420^2)$

$g = 21.5 m/s^2$

now we know that the acceleration due to gravity is given as

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

so we will have

$21.5 = \frac{(6.67 \times 10^{-11})M}{(7.40\times 10^7)^2}$

$M = 1.77 \times 10^{27} kg$

Answer 2

Acceleration caused due to the force of attraction is known as the acceleration due to gravity. The mass of the planet is 1.77x10²⁷ kg.

What is the acceleration due to gravity?

The acceleration due to gravity is the acceleration caused due to the force of attraction.

We know that the height from which the small rock is dropped is 1.90 m, while the time taken by the rock to touch the ground is 0.420 s, therefore, according to the second law of motion we can write the acceleration of the rock as,

$S= \dfrac{1}{2}gt^2\\\\1.90 = 0.5 \times g \times 0.42^2\\\\g = 21.542\rm\ m/s^2$

As the acceleration due to the gravity of the planet is known, therefore, the mass of the planet can be found using the formula for the acceleration,

$g=G\dfrac{m}{R^2}\\\\21.542 =6.67 \times 10^{-11}\dfrac{m}{(7.40\times 10^7)^2}\\\\M = 1.77 \times 10^{27}\rm\ kg$

Hence, the mass of the planet is 1.77x10²⁷ kg.

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