When a small object is launched from the surface of a fictitious planet with a speed of 52.0 m/s, its final speed when it is very far away from the planet is 32.0 m/s. Use this information to determine the escape speed of the planet.
squrt (53^2 - 32^2) = 42.249 m/s (Conservation of kinetic energy)
Well, let's see. If a small object is launched from the surface of a planet with a speed of 52.0 m/s, and its final speed when it's very far away is 32.0 m/s, I would say the planet needs to up its game in the launching department. It seems like the object was trying to escape, but the planet was like, "Nah, bruh, I'm gonna bring you back down." So, the escape speed of this planet would definitely be greater than 52.0 m/s, because otherwise, our small object would have successfully made its great escape. Keep dreaming, little object!
To determine the escape speed of the planet, we need to calculate the speed at which the object must be launched from the surface of the planet in order for it to reach a final speed of zero when very far away.
Escape speed is defined as the minimum speed at which an object must be launched from the surface of a planet in order to escape its gravitational pull.
Given that the initial speed of the object is 52.0 m/s and the final speed when very far away is 32.0 m/s, we can use the principle of conservation of mechanical energy to solve for the escape speed.
The conservation of mechanical energy states that the sum of the kinetic energy and potential energy of an object remains constant if only conservative forces are acting on it. In this case, the only conservative force acting on the object is gravity.
In the initial state, the object has kinetic energy (0.5 * m * v_initial^2) and potential energy (-G * M * m / r_initial), where m is the mass of the object, v_initial is the initial speed, G is the gravitational constant, M is the mass of the planet, and r_initial is the distance from the object to the center of the planet when launched.
In the final state, when the object is very far away, it has no kinetic energy (as its final speed is zero) and only potential energy (-G * M * m / r_final), where r_final is the distance from the object to the center of the planet when very far away.
Setting the initial and final states equal to each other:
0.5 * m * v_initial^2 - G * M * m / r_initial = -G * M * m / r_final
We can simplify this equation:
0.5 * v_initial^2 - G * M / r_initial = -G * M / r_final
Now, let's plug in the given values:
v_initial = 52.0 m/s
v_final = 0.0 m/s
Simplifying further:
(0.5 * 52.0^2) - G * M / r_initial = -G * M / r_final
(0.5 * 52.0^2) - G * M / r_initial = 0
Substituting the value of r_final as infinity (as we want to find the escape speed), the equation becomes:
(0.5 * 52.0^2) - G * M / r_initial = 0
Simplifying further:
1352 - G * M / r_initial = 0
Rearranging the equation to isolate the escape speed (v_escape):
v_escape = sqrt(2 * G * M / r_initial)
Therefore, the escape speed of the planet is given by the square root of (2 * G * M / r_initial) where G is the gravitational constant, M is the mass of the planet, and r_initial is the distance from the object to the center of the planet when launched.
To determine the escape speed of the planet, we need to find the minimum speed required for an object to completely escape the gravitational pull of the planet and not fall back down.
The escape speed can be calculated using the formula:
escape speed = sqrt(2 * gravitational constant * mass of the planet / radius of the planet)
In this case, we are given the initial speed (52.0 m/s) and the final speed (32.0 m/s) of the object when it is very far away from the planet. The final speed represents the speed the object has when the gravitational pull is negligible.
At very far distances from the planet, we can assume that the kinetic energy of the object is equal to the potential energy with respect to the planet. Thus, we can equate the kinetic and potential energies of the object at the initial and final states:
(1/2) * mass * initial speed^2 = - (G * mass of planet * object mass) / initial radius + (G * mass of planet * object mass) / final radius
Where:
- G is the gravitational constant (approximately 6.67430 × 10^-11 m^3 / (kg s^2))
- mass is the mass of the object being launched
- initial radius is the distance from the center of the planet to the launch point
- final radius is the distance from the center of the planet to the point where the object is very far away
Since we are considering the same object being launched at different distances from the planet, the mass of the object can be canceled out.
Simplifying the above equation, we can isolate the escape speed as follows:
(1/2) * initial speed^2 = G * mass of planet * (1 / initial radius - 1 / final radius)
Now, we can substitute the given values into the equation:
(1/2) * (52.0 m/s)^2 = (6.67430 × 10^-11 m^3 / (kg s^2)) * mass of planet * (1 / initial radius - 1 / final radius)
We don't have the mass of the planet or the radii, but we can still solve for the escape speed by rearranging the equation:
escape speed = sqrt(2 * (6.67430 × 10^-11 m^3 / (kg s^2)) * mass of planet * (1 / initial radius - 1 / final radius))
Therefore, the escape speed of the planet can be determined by plugging in the appropriate values for the mass of the planet, initial radius, and final radius into the above equation.