IBDP Physics- D.1 Gravitational fields- IB Style Questions For HL Paper 1A -FA 2025
▶️ Answer/Explanation
Gravitational field strength g = GM/R²
For Earth: g = GM_E/R_E²
For Planet P: g_P = GM_P/R_P² = G(M_E/18)/(R_E/3)² = (G M_E/18)/(R_E²/9) = (G M_E/R_E²) × (9/18) = g × 1/2
✅ Answer: (B) g/2
▶️ Answer / Explanation
The satellite is initially in a circular orbit at a distance \( r = R + \dfrac{R}{4} = \dfrac{5R}{4} \) from the centre of Earth.
The total mechanical energy of a satellite in circular orbit is:
\( E = -\dfrac{GMm}{2r} \)
Substituting \( r = \dfrac{5R}{4} \):
\( E = -\dfrac{GMm}{2 \times \dfrac{5R}{4}} = -\dfrac{2GMm}{5R} \)
At infinity, both the gravitational potential energy and kinetic energy are zero. Therefore, the energy required to move the satellite to infinity is:
\( E_{\text{required}} = \dfrac{2GMm}{5R} \)
The escape speed from the surface of Earth is given by:
\( v_{\mathrm{esc}} = \sqrt{\dfrac{2GM}{R}} \)
Squaring and substituting:
\( GM = \dfrac{v_{\mathrm{esc}}^{2} R}{2} \)
Hence,
\( E_{\text{required}} = \dfrac{2}{5} m \times \dfrac{v_{\mathrm{esc}}^{2} R}{2R} = \dfrac{m v_{\mathrm{esc}}^{2}}{5} \)
✅ Answer: (A)
▶️ Answer / Explanation
The gravitational field strength due to a point mass is \( g = \dfrac{GM}{r^{2}} \).
At X, the field due to mass P is \( g_P = \dfrac{Gm}{d^{2}} \).
The field due to mass Q is \( g_Q = \dfrac{G(2m)}{(2d)^{2}} = \dfrac{Gm}{2d^{2}} \).
These fields act in opposite directions, so the net gravitational field strength at X is
\( g_X = g_P – g_Q = \dfrac{Gm}{2d^{2}} \).
The gravitational potential due to a point mass is \( V = -\dfrac{GM}{r} \).
Potential at X due to P is \( V_P = -\dfrac{Gm}{d} \), and due to Q is \( V_Q = -\dfrac{Gm}{d} \).
Hence, the net gravitational potential at X is \( V_X = -\dfrac{2Gm}{d} \).
✅ Answer: D