Answer:
A) Object A is 3.25 times hotter.
B) Object A radiates 111.6 times more energy per unit of area.
Explanation:
Wiens's law states that there is an inverse relationship between the wavelength in which there is a peak in the emission of a black body and its temperature, mathematically,
,
where 
is the temperature in kelvins and, 
is the wavelenght (in meters) where the emission is in its peak.
From here, if we solve Wien's law for the temperature we get
.
Now, we can easily compute the temperatures.
For object A:
.
For object B:
From this, we get that
,
which means that object A is 3.25 times hotter.
Stefan's Law states that a black body emits thermal radiation with power proportional to the fourth power of its temperature.
This is
,
where 
is call the Stefan-Boltzmann constant.
From this, power can be easily compute:
,
and we can notice that
,
which means that object A radiates 111.6 time more energy per unit of area.
The forces of gravity between two objects are inversely proportional to
the square of the distance between them. So reducing the distance
by 1/2 means increasing the gravitational force by 2² = 4 times.
The 1 million newtons becomes 4 million newtons.
Note that this does NOT mean the satellite's altitude above the surface.
When you're calculating gravitational forces, it's the distance between
the centers of the objects. So the question is a meaningful exercise
only if we use the distance between the satellite and the planet's center.
Answer:
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<span>Neo and Morpheus's masses have gained a velocity (not equal to zero) which means their momentum is now based on gravity and friction alone.</span>
Answer:
b) Betelgeuse would be 
times brighter than Sirius
c) Since Betelgeuse brightness from Earth compared to the Sun is 
the statement saying that it would be like a second Sun is incorrect
Explanation:
The start brightness is related to it luminosity thought the following equation:
(1)
where 
is the brightness, 
is the star luminosity and 
, the distance from the star to the point where the brightness is calculated (measured). Thus:
b) 
and 
where 
is the Sun luminosity (
) but we don't need to know this value for solving the problem. 
is light years.
Finding the ratio between the two brightness we get:
c) we can do the same as in b) but we need to know the distance from the Sun to the Earth, which is 
. Then
Notice that since the star luminosities are given with respect to the Sun luminosity we don't need to use any value a simple states the Sun luminosity as the unit, i.e 1. From this result, it is clear that when Betelgeuse explodes it won't be like having a second Sun, it brightness will be 5 orders of magnitude smaller that our Sun brightness.
