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If the geostationary satellite is stationary in the observer's frame on earth, (which it actually is), then it's relative velocity will be zero. The problem lies in writing the expression for relative velocity correctly. The observer's frame is a rotating frame, which is to say that the imaginary coordinate axes fixed to him are rotating. So relative ...


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It is important to keep in the back of your mind that $\theta$ is really a function of time $\theta(t)$ and that using this function, one can find the velocity (not speed) of particle $A$ at time $t$: $v(\cos \theta(t), \sin \theta(t))$. Ultimately, we can plug and chug with this velocity vector and get a differential equation of $\theta$; however, this ...


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If your two objects are not subject to external forces, then gravity would accelerate them toward each other. They can do this with a fixed separation only if they are orbiting each other with both velocity vectors changing direction as a function of time.


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Turning my comment into an answer ... The velocity can be non-zero because velocity is represented by a vector, so the vector changes when the direction changes, even if the magnitude ("speed" in this case) remains constant. For example, consider the case of two objects in circular orbits around the barycenter. If I am riding on top of one of them ...


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The answer lies in your question itself. The two particles neither approach each other i.e. their separation isn't decreasing, nor are they moving apart i.e. their separation isn't increasing In other words with respect to both the particles, the separation between them is not changing. Relative velocity is given as the rate of change in relative separation (...


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Time is independent of frames. So it does not depend on your frame of reference whereas distance, velocity, acceleration, force etc. might change.


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Muzzle speed of the bullet means velocity of the bullet with respect to the velocity of the gun.


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What correction does the calculation from non-inertial frame need? The calculation from the non-inertial frame does not need any correction. It is an accurate calculation. Note that the mechanism that launched the object still used 139 J and the KE changed by 154 J, so in this non-inertial frame energy is not conserved. Energy is not necessarily conserved ...


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In noninertial case you computed kinetic energy at two different frames. In 1D case, when you boost the frame by $\Delta V$ in x direction the kinetic energy changes: $$E_k=\sum_i \frac{1}{2}m_iv_i^2 \rightarrow E'_k=\sum_i \frac{1}{2}m_i(v_i-\Delta V)^2=E_k-\frac{1}{2}\Delta V\sum_i m_i v_i+ \frac{1}{2}M\Delta V^2,$$ where $M=\sum_i m_i.$ In your case, the ...


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