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

Question

Which is a statement of one of Kepler’s laws of orbital motion?

(A) The square of the planet’s orbital period is proportional to the cube of the length of the semi-major axis of its orbit.
(B) A line segment joining a planet and the Sun sweeps out equal arc lengths during equal intervals of time.
(C) A planet’s orbital period is proportional to the cube of the length of the semi-major axis of its orbit.
(D) The orbit of a planet is an ellipse with the Sun positioned at the centre.

▶️ Answer/Explanation

Correct Answer: \( \boxed{\mathrm{A}} \)

Kepler’s third law states that the square of the orbital period is proportional to the cube of the semi-major axis:

\(T^2\propto a^3\)

where \(T\) is the orbital period and \(a\) is the semi-major axis of the orbit.

Option B is incorrect because Kepler’s second law refers to equal areas, not equal arc lengths, being swept out in equal intervals of time.

Option C incorrectly states \(T\propto a^3\), rather than \(T^2\propto a^3\).

Option D is incorrect because the Sun is located at one focus of the elliptical orbit, not at its centre.

Hence, the correct answer is \( \boxed{\mathrm{A}} \).

Question

Isolated planets X and Y have masses \(M_X\) and \(M_Y\) respectively and are separated by a distance \(R\).

A point P is located at a distance \(\dfrac{R}{3}\) from planet X, as shown. The gravitational field strength at P is zero.

What is \(\dfrac{M_Y}{M_X}\)?

(A) \(2\)
(B) \(3\)
(C) \(4\)
(D) \(9\)

▶️ Answer/Explanation

Correct Answer: \( \boxed{\mathrm{C}} \)

At P, the gravitational fields due to X and Y are equal in magnitude and opposite in direction.

The gravitational field strength due to a mass \(M\) at distance \(r\) is

\(g=\dfrac{GM}{r^2}\)

The distance from X to P is \(\dfrac{R}{3}\), while the distance from P to Y is

\(R-\dfrac{R}{3}=\dfrac{2R}{3}\)

Since the resultant gravitational field at P is zero,

\(\dfrac{GM_X}{(R/3)^2}=\dfrac{GM_Y}{(2R/3)^2}\)

Cancelling \(G\) and \(R^2\),

\(\dfrac{M_X}{1/9}=\dfrac{M_Y}{4/9}\)

Therefore,

\(9M_X=\dfrac{9}{4}M_Y\)

\(\dfrac{M_Y}{M_X}=4\)

Hence, the correct answer is \( \boxed{\mathrm{C}} \).

Question

Two isolated point masses, P of mass \(m\) and Q of mass \(2m\), are separated by a distance \(3d\). X is a point a distance \(d\) from P and \(2d\) from Q.

What is the net gravitational field strength at X and the net gravitational potential at X?

Option	Net gravitational field strength at X	Net gravitational potential at X
A	\( \dfrac{Gm}{d^{2}} \)	\( 0 \)
B	\( \dfrac{Gm}{d^{2}} \)	\( -\dfrac{2Gm}{d} \)
C	\( \dfrac{Gm}{2d^{2}} \)	\( 0 \)
D	\( \dfrac{Gm}{2d^{2}} \)	\( -\dfrac{2Gm}{d} \)

▶️ Answer / Explanation

Detailed solution

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

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

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