The available options are: (found the complete text on internet)
A- at a distance less than r
B- at a distance equal to r
C- at a distance greater than r
Solution:
The correct answer is C) at a distance greater than r.
In fact, the gravitational attraction between the satellite and the Earth provides the centripetal force that keeps the satellite in circular orbit, so we can write
[tex]G \frac{Mm}{r^2}=m \frac{v^2}{r} [/tex]
where the term on the left is the gravitational force, while the term on the right is the centripetal force, and where
G is the gravitational constant
M is the Earth mass
m is the satellite mass
r is the distance of the satellite from the Earth's center
v is the satellite speed
Re-arranging the equation, we get
[tex]r= \frac{GM}{v^2} [/tex]
and we see from this formula that, if the second satellite has a speed less than the speed v of the first satellite, it means that the denominator of the fraction is smaller, and so r is larger for the second satellite.