Find the speed of a satellite at a distance R from Earth

In summary, the problem statement is asking for the velocity of a satellite at a distance of two from the center of the Earth. However, the equation that is given does not correctly express the centripetal force required for a circular trajectory. The answer is found by solving for v.
  • #36
Ok. My answer is like Ep + Ek
While the formula i know for Ep is -GMm/r
r = 2R
Ek as usual is 1/2mv^2
V = ##\sqrt{ gmR/2}##

Ep + Ek
-GMm/2R + mgR/4 = -mgR/2 + mgR/4 = -mgR/4
Is that it?
 
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  • #37
PeterO said:
NO. The Mechanical Energy is the sum of the Gravitational Potential Energy plus the Kinetic Energy. The Kinetic Energy may be positive, but the Mechanical Energy value can still be negative, provided the Potential Energy is negative, and larger in size than the Kinetic Energy.
PeterO said:
NO. The Mechanical Energy is the sum of the Gravitational Potential Energy plus the Kinetic Energy. The Kinetic Energy may be positive, but the Mechanical Energy value can still be negative, provided the Potential Energy is negative, and larger in size than the Kinetic Energy.
I see
 
  • #38
Helly123 said:
Ok. My answer is like Ep + Ek
While the formula i know for Ep is -GMm/r
r = 2R
Ek as usual is 1/2mv^2
V = ##\sqrt{ gmR/2}##

Ep + Ek
-GMm/2R + mgR/4 = -mgR/2 + mgR/4 = -mgR/4
Is that it?
Yes.
 
  • #39
ehild said:
Yes.
Ok
 
  • #40
The potential energy is negative and twice the magnitude of the kinetic energy (which is positive), thus the total energy is negative. So work is required to escape the gravitational well of the Earth no matter where you are.
 
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