If an astronaut dropped a small rock near the surface of Pluto, how far (in m) would the rock fall in 1.1 s? (The mass of Pluto is 2.500×10-3 the mass of earth and the radius of Pluto is 0.18 of the earth's radius.)

Since g=GM/r²

G=gravitational constant
M=mass of planet
r=radius of planet
Calculate g (by relative proportions)
on Pluto and use new value of g for the free fall equation.

Well, if an astronaut dropped a small rock near the surface of Pluto, it would be quite a cosmic game of "catch" between the rock and Pluto's gravity. So, let me humor you with some calculations.

Pluto's gravity is weaker compared to Earth's due to its smaller mass. But don't worry, Pluto still knows how to keep things interesting! Let's do some math.

The acceleration due to gravity on Pluto can be found using the formula:

g_pluto = (G * m_pluto) / r_pluto^2

Where G is the gravitational constant, m_pluto is the mass of Pluto, and r_pluto is the radius of Pluto.

Now, since you mentioned the mass of Pluto is 2.500×10^-3 the mass of Earth, we can say:

m_pluto = 2.500×10^-3 * m_earth

Now, the radius of Pluto is 0.18 times the radius of Earth, so:

r_pluto = 0.18 * r_earth

Substituting these values into the gravity equation, we get:

g_pluto = (G * (2.500×10^-3 * m_earth)) / (0.18 * r_earth)^2

Now, plug in the known values for G, m_earth, and r_earth, and calculate g_pluto.

Once you have g_pluto, you can use the equation for distance traveled during free fall:

d = (1/2) * g * t^2

Where d represents the distance fallen, g represents the acceleration due to gravity, and t represents the time in seconds.

Now you can calculate the distance the rock would fall in 1.1 seconds on Pluto! But remember, this is all just imaginary because it's highly unlikely that astronauts are dropping rocks near the surface of Pluto. But hey, we can always use our imagination, right? Happy falling!

To determine how far the rock would fall in 1.1 seconds near the surface of Pluto, we can use the equation for free-fall acceleration:

d = (1/2) * g * t^2

Where:
d = distance fallen
g = acceleration due to gravity
t = time

First, let's find the acceleration due to gravity on Pluto. Since the mass of Pluto is given as 2.500×10^(-3) the mass of Earth, we can calculate the acceleration due to gravity on Pluto compared to Earth using the gravitational force equation:

F = (G * m * M) / r^2

Where:
F = gravitational force
G = gravitational constant ≈ 6.67430 x 10^-11 N(m/kg)^2
m = mass of the rock
M = mass of Pluto
r = radius of Pluto

The ratio of the force on Pluto to the force on Earth is given by:

(F_pluto / F_earth) = (g_pluto / g_earth) = (M_pluto / M_earth) * (r_earth^2 / r_pluto^2)

Substituting the given values:
(M_pluto / M_earth) = 2.500×10^(-3)
(r_earth^2 / r_pluto^2) = (1 / 0.18)^2 = 32.72

Therefore:
(g_pluto / g_earth) = 2.500×10^(-3) * 32.72

Now, let's find the acceleration due to gravity on Earth, which is approximately 9.8 m/s^2.

Therefore:
g_pluto = 9.8 m/s^2 * 2.500×10^(-3) * 32.72

Now, substituting the value of g_pluto into the equation for the distance fallen:

d = (1/2) * g_pluto * t^2

d = (1/2) * (9.8 m/s^2 * 2.500×10^(-3) * 32.72) * (1.1 s)^2

By calculating the expression above, you will find the distance the rock would fall in 1.1 seconds near the surface of Pluto.

To calculate the distance that the rock would fall near the surface of Pluto in 1.1 seconds, we need to first determine the acceleration due to gravity on Pluto.

The formula for the acceleration due to gravity is:

\[ g = \frac{{G \cdot M}}{{r^2}} \]

Here, g represents the acceleration due to gravity, G is the gravitational constant (which is approximately \(6.67430 \times 10^{-11} \, \text{Nm}^2/\text{kg}^2\)), M is the mass of Pluto, and r is the radius of Pluto.

Since we know the mass of Pluto is \(2.500 \times 10^{-3}\) times the mass of Earth, and the radius of Pluto is \(0.18\) times the radius of Earth, we can substitute these values into the equation.

The mass of Earth is approximately \(5.972 \times 10^{24} \, \text{kg}\), and the radius of Earth is about \(6.371 \times 10^{6} \, \text{m}\). So we have:

\[ M = 2.500 \times 10^{-3} \times 5.972 \times 10^{24} \, \text{kg} \]
\[ r = 0.18 \times 6.371 \times 10^{6} \, \text{m} \]

Let's calculate these values:

\[ M = 1.493 \times 10^{21} \, \text{kg} \]
\[ r = 1.1478 \times 10^{6} \, \text{m} \]

Now, we can substitute these values into the formula to calculate the acceleration due to gravity on Pluto:

\[ g = \frac{{6.67430 \times 10^{-11} \, \text{Nm}^2/\text{kg}^2 \times 1.493 \times 10^{21} \, \text{kg}}}{{(1.1478 \times 10^{6} \, \text{m})^2}} \]

Calculating this expression gives us:

\[ g \approx 0.592 \, \text{m/s}^2 \]

This means that the acceleration due to gravity on Pluto is approximately \(0.592 \, \text{m/s}^2\).

Now, to find the distance that the rock would fall in 1.1 seconds, we can use the equation of motion:

\[ d = \frac{1}{2} g t^2 \]

Where d represents the distance, g is the acceleration due to gravity, and t is the time.

Substituting the values we know:

\[ d = \frac{1}{2} \times 0.592 \, \text{m/s}^2 \times (1.1 \, \text{s})^2 \]

Calculating this expression gives us:

\[ d \approx 0.374 \, \text{m} \]

So, the rock would fall approximately 0.374 meters near the surface of Pluto in 1.1 seconds.