Answer:
Half
Explanation:
Given that:
- radial distance of satellite from the earth,

Now, if the satellite is moved to a distance 
<u>We have the mathematical expression for the potential energy fue to gravitational field as:</u>
...................(1)
where:

M = mass of earth
m = mass of satellite
R = radial distance of satellite
<u>Now from eq. (1) initially we have:</u>

<u>after the satellite is moved, we have:</u>



which is half of the initial condition.
Answer: 5,640 s (94 minutes)
Explanation:
the tangential speed of the HST is given by
(1)
where
is the length of the orbit
r is the radius of the orbit
T is the orbital period
In our problem, we know the tangential speed:
. The radius of the orbit is the sum of the Earth's radius and the distance of the HST above Earth's surface:

So, we can re-arrange equation (1) to find the orbital period:

Dividing by 60, we get that this time corresponds to 94 minutes.
Answer:
What are we supposed to find, if it is kinetic energy then this is the solution.
K.E=1/2mv^2
K.E= kinetic energy
M=mass
V=velocity
K.E =0.5*55*0.6^2
K.E=9.9J
Explanation:
Homie I don’t know either♀️Drop out of schl ig ;-;