R S ( M ) = 2 G M c 2 , where G is the gravitational constant and c is the speed of light. It is okay if you do not follow the d
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Answer:
Speed of the spacecraft right before the collision: .
Assumption: the earth is exactly spherical with a uniform density.
Explanation:
This question could be solved using the conservation of energy.
The mechanical energy of this spacecraft is the sum of:
- the kinetic energy of this spacecraft, and
- the (gravitational) potential energy of this spacecraft.
Let denote the mass of this spacecraft. At a distance of
from the center of the earth (with mass
), the gravitational potential energy (
) of this spacecraft would be:
.
Initially, (the denominator of this fraction) is infinitely large. Therefore, the initial value of
will be infinitely close to zero.
On the other hand, the question states that the initial kinetic energy () of this spacecraft is also zero. Therefore, the initial mechanical energy of this spacecraft would be zero.
Right before the collision, the spacecraft would be very close to the surface of the earth. The distance between the spacecraft and the center of the earth would be approximately equal to
, the radius of the earth.
The of the spacecraft at that moment would be:
.
Subtract this value from zero to find the loss in the of this spacecraft:
Assume that gravitational pull is the only force on the spacecraft. The size of the loss in the of this spacecraft would be equal to the size of the gain in its
.
Therefore, right before collision, the of this spacecraft would be:
.
On the other hand, let denote the speed of this spacecraft. The following equation that relates
and
to
:
.
Rearrange this equation to find an equation for :
.
It is already found that right before the collision, . Make use of this equation to find
at that moment:
.
Answer:
Explanation:
= Initial pressure =
= Initial volume
= Final volume =
Temperature is the same in the initial and final state
From the ideal gas law we have
The final pressure of the system is .