A body is dropped from a height of 10 meters on Earth and hits the ground with a certain velocity. The body is then taken to the moon, which has a gravitational constant that is 1/6th the value as compared to Earth. It is dropped from the same height, hitting the ground with a certain velocity. What is the ratio of the object's velocity on the moon to that of the earth?
To find the ratio of the object's velocity on the moon to that on Earth, we can use the concept of gravitational potential energy and kinetic energy.
On Earth:
When the body is dropped from a height of 10 meters on Earth, it initially has potential energy due to its height, which will be converted into kinetic energy as it falls. At the moment it hits the ground, all the potential energy is converted into kinetic energy.
On the Moon:
The gravitational constant on the moon is 1/6th the value of Earth's, which means the moon's gravitational acceleration is also 1/6th that of Earth's. Hence, the object will experience a weaker gravitational pull on the moon compared to Earth.
To calculate the ratio of the object's velocity on the moon to that on Earth, we can use the conservation of mechanical energy, which says that the sum of potential energy and kinetic energy remains constant in the absence of external forces.
Let's denote the object's velocity on Earth as Ve and on the moon as Vm.
On Earth:
The potential energy (PE) of the body is given by PE = mgh, where m is the mass, g is the acceleration due to gravity on Earth, and h is the height. The initial potential energy at a height of 10 meters is PEe = m * 9.8 * 10 Joules.
At the moment it hits the ground, all the potential energy is converted into kinetic energy (KE). Therefore, KEe = PEe, which can be written as KEe = (1/2)mv^2, where v is the velocity of the body on Earth.
On the Moon:
Since the gravitational constant on the moon is 1/6th of Earth's, the gravitational acceleration on the moon (gm) will be equal to (1/6) * 9.8 m/s^2.
The potential energy of the body on the moon (PEm) is given by PEm = m * gm * h. Substituting the values, PEm = m * (1/6) * 9.8 * 10 Joules.
At the moment it hits the ground, all the potential energy is converted into kinetic energy (KEm). Therefore, KEm = PEm, which can be written as KEm = (1/2)mv^2, where v is the velocity of the body on the moon.
Now, we can calculate the ratio of the object's velocity on the moon to that on Earth:
Vm/Ve = √(KEm/KEe) = √(PEm/PEe) = √((m * (1/6) * 9.8 * 10) / (m * 9.8 * 10))
Simplifying the equation:
Vm/Ve = √((1/6) / 1)
Vm/Ve = √(1/6)
Vm/Ve = 1/√6
Therefore, the ratio of the object's velocity on the moon to that on Earth is 1/√6.