Step 1: Geometry of the Loop

The problem states the loop height is not altered, so the vertical semi-axis is $R$. We are also given that the “semi major axis is equal to R”. This implies the vertical axis is the major axis ($a=R$), and the horizontal axis is the minor axis ($b < R$).

The loop is a “tall” ellipse.

Step 2: Radius of Curvature

The minimum velocity required to complete the loop depends on the radius of curvature $\rho$ at the highest point.

$$\rho = \frac{(\text{horizontal axis})^2}{\text{vertical axis}} = \frac{b^2}{R}$$

Since the horizontal axis $b < R$, the radius of curvature $\rho < R$. The turn is "sharper" than a circle.

Step 3: Energy Calculation

For a general loop of height $2R$ and top curvature $\rho$, the minimum release height $H$ is:

$$H = 2R + \frac{\rho}{2}$$

Comparing the two cases:

• Circle (P): $\rho = R \implies H_P = 2.5R$
• Ellipse (Q): $\rho < R \implies H_Q < 2.5R$