The planet Krypton has a mass of 9*10^ 23 kg and radius of 1.2 * 10 ^ 6 * m What is the acceleration of an object in free fall near the surface of Krypton? The gravita tional constant is 6.6726*10^ -11 N* m^ 2 /kg^ 2 Answer in units of m/s^ 2
GM/r^2 = 6.6726*10^-11 * 9*10^23 / (1.2*10^6)^2 = 41.7 m/s^2
To calculate the acceleration of an object in free fall near the surface of Krypton, we can use Newton's law of universal gravitation.
The law states that the force of gravity between two objects is directly proportional to the product of their masses and inversely proportional to the square of the distance between their centers. Mathematically, it can be expressed as:
F = (G * m1 * m2) / r^2
Where:
- F is the force of gravity between the two objects.
- G is the gravitational constant (6.6726 * 10^-11 N * m^2 / kg^2).
- m1 and m2 are the masses of the two objects.
- r is the distance between the centers of the two objects.
In this case, we are interested in the acceleration caused by the gravitational force near the surface of Krypton. Near the surface of a planet, the gravitational force is approximately equal to the weight of an object. So, we can rearrange the equation to solve for acceleration:
F = m * a (force = mass * acceleration)
Equating the equations, we have:
m * a = (G * m1 * m2) / r^2
Since we want to find the acceleration, we can solve for 'a':
a = (G * m1 * m2) / (m * r^2)
In this case, we are considering an object falling near the surface of Krypton. So, m1 and r will be the mass and radius of the planet, and m2 will be the mass of the falling object.
Substituting the values given:
- m1 = mass of Krypton = 9 * 10^23 kg
- m = mass of the falling object (not provided)
- r = radius of Krypton = 1.2 * 10^6 m
- G = gravitational constant = 6.6726 * 10^-11 N * m^2 / kg^2
Now we can compute the acceleration by substituting the values into the equation and solving it:
a = (G * m1 * m2) / (m * r^2)
= (6.6726 * 10^-11 N * m^2 / kg^2 * 9 * 10^23 kg * m2) / (m * (1.2 * 10^6 m)^2)
= (6.6726 * 9 * 10^23 m2 / (1.2 * 10^6 m)^2) * (N * m^2 / kg^2 / m)
= (6.6726 * 9 * 10^23) / (1.2 * 10^6)^2
= (6.6726 * 9 * 10^23) / (1.44 * 10^12)
= 4.5 * 10^11
Therefore, the acceleration of an object in free fall near the surface of Krypton is approximately 4.5 * 10^11 m/s^2.