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IBDP Physics- D.1 Gravitational fields- IB Style Questions For SL Paper 1A -FA 2025

IBDP Physics SL Paper 1 -FA 2025- All Chapters

Question

Planet P has a diameter equal to one third of the Earth’s diameter. The Earth’s mass is \(18\) times greater than that of planet P. Given that the gravitational field strength at the Earth’s surface is \(g\), determine the gravitational field strength at the surface of planet P.
(A) \(\frac{g}{6}\)
(B) \(\frac{g}{2}\)
(C) \(2g\)
(D) \(6g\)
▶️ Answer/Explanation
Detailed solution

1. Gravitational Field Strength Formula:
\(g = \frac{GM}{r^2}\).

2. Express Planet Parameters:

  • \(M_P = \frac{1}{18} M_E\)
  • \(r_P = \frac{1}{3} r_E\) (diameter ratio equals radius ratio)

3. Calculate \(g_P\):
\(g_P = \frac{G M_P}{r_P^2} = \frac{G (\frac{1}{18} M_E)}{(\frac{1}{3} r_E)^2}\)
\(g_P = \frac{\frac{1}{18}}{\frac{1}{9}} \frac{G M_E}{r_E^2}\)
\(g_P = \frac{9}{18} g = \frac{1}{2} g\).
✅ Answer: (B)

Question

A ball is projected at an angle to the horizontal on Earth, reaching a maximum height \(H\) and a maximum range \(R\). The same ball is projected at the same angle and speed on a planet where the acceleration due to gravity is three times that on Earth. Air resistance is negligible.
What are the maximum range and the maximum height reached on that planet?
\( \displaystyle \begin{array}{c|c|c} \textbf{Option} & \textbf{Maximum range} & \textbf{Maximum height reached}\\ \hline \text{A} & \dfrac{R}{3} & \dfrac{H}{3}\\ \hline \text{B} & R & \dfrac{H}{3}\\ \hline \text{C} & R & \dfrac{H}{9}\\ \hline \text{D} & \dfrac{R}{3} & \dfrac{H}{9} \end{array} \)
▶️ Answer/Explanation

✅ Answer: (A)

On level ground (same launch speed \(V\) and angle \(\theta\)):
Maximum range: \(R = \dfrac{V^{2}\sin(2\theta)}{g}\)
Maximum height: \(H = \dfrac{V^{2}\sin^{2}(\theta)}{2g}\)

Both \(R\) and \(H\) are inversely proportional to \(g\).
On the planet: \(g_p = 3g\).

So,
\(R_p = \dfrac{V^{2}\sin(2\theta)}{3g} = \dfrac{R}{3}\)
\(H_p = \dfrac{V^{2}\sin^{2}(\theta)}{2(3g)} = \dfrac{H}{3}\)

Therefore the maximum range is \(\dfrac{R}{3}\) and the maximum height is \(\dfrac{H}{3}\).

Question

The radius of the Earth is \(R\). A satellite is launched to a height \(h=\frac{R}{4}\) above the Earth’s surface. What is \(\frac{\text{gravitational force on satellite at the surface}}{\text{gravitational force on satellite at height }h}\)?
(A) \(\frac{4}{5}\)
(B) \(\frac{16}{25}\)
(C) \(\frac{25}{16}\)
(D) \(\frac{5}{4}\)
▶️ Answer/Explanation
Detailed solution

Gravitational force varies with distance from Earth’s centre as \(F \propto \frac{1}{r^2}\).

At the surface: \(r = R\), so \(F_{\text{surface}} \propto \frac{1}{R^2}\).

At height \(h=\frac{R}{4}\):
\(r = R+h = R+\frac{R}{4}=\frac{5R}{4}\), so \(F_{h} \propto \frac{1}{\left(\frac{5R}{4}\right)^2}\).

Therefore, \[ \frac{F_{\text{surface}}}{F_h} = \frac{\frac{1}{R^2}}{\frac{1}{\left(\frac{5R}{4}\right)^2}} = \left(\frac{5R}{4}\right)^2 \frac{1}{R^2} = \left(\frac{5}{4}\right)^2 = \frac{25}{16}. \]

✅ Answer: (C)

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