Question

111. When a meteorite strikes Earth, the meteorite or its pieces remain on or near Earth's surface. Suppose a giant object from space enters the solar system and collides head-on with Earth as Earth travels in its path around the Sun.

If the object's speed is twice Earth's speed, how massive must it be to reduce Earth's orbital velocity by even a hundredth of a percent? Annotate your calculation with a description of your approach.​

Answer 1

Step-by-step explanation:

To calculate the mass of an object that would reduce Earth's orbital velocity by even a hundredth of a percent, we need to use the conservation of momentum principle. According to this principle, the total momentum of a system remains constant if no external forces act on it.

In this scenario, we can consider the Earth and the object as a closed system. The initial momentum of the system is Earth's momentum, and the final momentum is Earth's momentum after the collision.

Let's assume the mass of the Earth is M_E and the initial velocity of Earth is V_E. After the collision with the object, the final velocity of Earth is V_E' . The mass of the object is M_O, and its velocity is V_O (twice Earth's velocity)

The conservation of momentum principle states that:

M_E * V_E = M_E * V_E' + M_O * V_O

To reduce Earth's orbital velocity by even a hundredth of a percent, we can assume that V_E' = V_E - (0.0001 * V_E)

Therefore,

M_E * V_E = M_E * (V_E - (0.0001 * V_E)) + M_O * (2 * V_E)

Rearranging this equation we can find the mass of the object M_O

M_O = M_E * (V_E / (2 * V_E - 0.0001 * V_E))

To get the exact mass of the object, we need to know the mass of the Earth and the velocity of Earth, which are not provided in the question. However, this calculation will give an approximate mass that the object would need to have in order to reduce Earth's orbital velocity by even a hundredth of a percent.