Question

If the horizontal velocity given to a satellite is greater than critical velocity but less than the escape velocity at the height, then the satellite will

• A. be lost in space
• B. falls on the earth along parabolic path
• C. revolve in circular orbit
• D. revolve in elliptical orbit

Hint:

Think of circular velocity as a single perfect balance point. Any variation in speed that doesn't reach the absolute escape threshold ($v_e$) will automatically default the path into an ellipse. Extra speed expands the orbit into an ellipse outside the circle, while lesser speed shrinks it inside.

Answer 1

The Correct Option is D

Step 1: Anchor on total mechanical energy.
Instead of memorising the trajectory chart, decide the orbit shape from the sign of the total energy $E = K + U$, where $U = -\dfrac{GMm}{R+h}$ is negative.
Step 2: Escape velocity sets the zero-energy mark.
At $v = v_e$, kinetic energy exactly cancels the negative potential energy, giving $E = 0$ (a parabola). Above $v_e$, $E>0$ (a hyperbola, unbound).
Step 3: Critical velocity sets the circular case.
At $v = v_c$, gravity supplies exactly the centripetal need, giving a circle.
Step 4: Place our satellite.
We are told $v_c < v < v_e$. Since $v < v_e$, total energy is still negative, so the orbit stays closed and bound.
Step 5: Rule out the circle.
Since $v > v_c$, the speed exceeds the perfect-circle value, so the path is not circular.
Step 6: The only bound non-circular orbit and conclude.
A bound orbit that is not circular must be an ellipse (Kepler's first law). \[ \boxed{\text{Elliptical orbit}} \]