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: Understanding the Question:
The question concerns the orbital trajectory of a satellite launched horizontally from a specific altitude above Earth with an initial velocity $v_h$.
We are given that this launching velocity lies strictly between two boundary values: the critical circular velocity $v_c$ and the escape velocity $v_e$ ($v_c < v_h < v_e$).

Step 2: Key Formula or Approach:
The trajectory of an object in a central gravitational field depends on its launch velocity relative to standard orbital benchmarks:
1. If $v_h < v_c$, the satellite moves in an elliptical path with the launch point as its apogee, eventually entering the atmosphere or crashing.
2. If $v_h = v_c$, the satellite travels in a perfectly stable circular orbit.
3. If $v_c < v_h < v_e$, the satellite forms a stable bound elliptical orbit, where the launch point acts as the perigee (closest point to Earth).
4. If $v_h = v_e$, the path becomes an open parabolic escape trajectory.
5. If $v_h > v_e$, the path is a hyperbolic escape trajectory.

Step 3: Detailed Explanation:
Critical velocity ($v_c = \sqrt{\frac{GM}{R+h}}$) is the exact speed required to maintain a perfect circle where gravitational pull matches the required centripetal force.
Escape velocity ($v_e = \sqrt{\frac{2GM}{R+h}}$) is the speed at which total mechanical energy becomes zero, allowing an object to break free completely from Earth's gravity.
When the horizontal velocity $v_h$ exceeds $v_c$, the gravitational pull at that instantaneous radius is no longer strong enough to bend the path into a perfect circle. The satellite begins to move outward away from the Earth, climbing in altitude.
As it climbs, gravity slows it down until its velocity drops below the local critical velocity, at which point it rounds out and starts falling back toward Earth. This cyclic rise and fall forms a stable closed ellipse.
Since $v_h < v_e$, its total mechanical energy remains negative, meaning it cannot escape into deep space and remains bound in an elliptical path.

Step 4: Final Answer:
The satellite will revolve in an elliptical orbit, which matches option (D).