Apply Gravitational Fields - IB Physics

A space station orbits Earth at an altitude where the gravitational field strength is $4.9 \text{ N kg}^{-1}$ . An astronaut in the station drops a tool, which appears to float beside her. From the perspective of an observer on Earth's surface (where $g = 9.8 \text{ N kg}^{-1}$ ), what is the acceleration of the tool?
Answer: $4.9 \text{ m s}^{-2}$ toward Earth's center This question tests your understanding of reference frames and gravitational acceleration in orbital mechanics. The key insight is recognizing that different observers measure different accelerations for the same object. From an Earth-based observer's perspective, the tool experiences gravitational acceleration equal to the local gravitational field strength at the space station's altitude. Since the gravitational field strength is $4.9 \text{ N kg}^{-1}$ at that altitude, the tool accelerates at $4.9 \text{ m s}^{-2}$ toward Earth's center. This acceleration keeps the tool in orbit alongside the space station. Looking at the incorrect options: Option A ( $0 \text{ m s}^{-2}$ ) reflects what the astronaut observes in her reference frame—the tool appears to float because both she and the tool are in free fall together. However, the question specifically asks for the Earth observer's perspective. Option B ( $9.8 \text{ m s}^{-2}$ ) incorrectly uses Earth's surface gravity, but gravitational acceleration decreases with distance from Earth's center. At the space station's altitude, gravity is weaker than at Earth's surface. Option D suggests tangential acceleration, but gravitational force always acts radially toward Earth's center, not tangentially. The tool appears weightless to the astronaut because both are accelerating at the same rate due to gravity—this is the essence of free fall in orbit. Study tip: When analyzing orbital motion problems, always clarify which reference frame you're using. Objects in orbit are simultaneously accelerating (from an external observer's view) and weightless (from their own reference frame).
A binary star system consists of two stars of masses $M$ and $2M$ separated by distance $d$ . At what distance from the star of mass $M$ is the gravitational field strength zero?
Answer: $\frac{d}{1 + \sqrt{2}}$ Let the point be at distance $x$ from mass $M$ and $(d-x)$ from mass $2M$ . For zero field: $\frac{GM}{x^2} = \frac{G(2M)}{(d-x)^2}$ . This gives $\frac{1}{x^2} = \frac{2}{(d-x)^2}$ , so $(d-x) = x\sqrt{2}$ . Solving: $d = x + x\sqrt{2} = x(1 + \sqrt{2})$ , therefore $x = \frac{d}{1 + \sqrt{2}}$ . Option A assumes equal division by mass ratio, B assumes midpoint, C involves incorrect square root manipulation.
The escape speed from the surface of a planet of mass M and radius R is v_esc. What is the escape speed from a planet with mass 4M and radius R/2?
Answer: 2√2 v_esc The escape speed is given by the formula v_esc = √(2GM/R). For the new planet, the mass M' = 4M and the radius R' = R/2. The new escape speed v'_esc = √(2G(4M)/(R/2)) = √(16GM/R) = √8 * √(2GM/R) = 2√2 v_esc.
A satellite is in a geostationary orbit. Which of the following quantities must be the same for the satellite and a point on the Earth's equator?
Answer: The angular velocity A geostationary orbit means the satellite remains above the same point on the equator. For this to happen, it must have the same orbital period as the Earth's rotation (24 hours) and orbit in the same direction. Having the same period means it must have the same angular velocity (ω = 2π/T). The centripetal acceleration, gravitational force, and orbital speed all depend on the orbital radius, which is much larger for the satellite than for a point on the equator, so these quantities will be different.
What is the ratio of the orbital speed of a satellite in a low Earth orbit (radius R ≈ R_Earth) to the escape speed from the Earth's surface?
Answer: 1/√2 The orbital speed is v_orb = √(GM/R). The escape speed is v_esc = √(2GM/R). The ratio v_orb / v_esc = √(GM/R) / √(2GM/R) = √[(GM/R) / (2GM/R)] = √(1/2) = 1/√2.
A satellite of mass m orbits a planet of mass M in a circle of radius r. What is the total energy of the satellite?
Answer: -GMm/(2r) The total energy E is the sum of kinetic energy (KE) and potential energy (PE). PE = -GMm/r. For a circular orbit, mv²/r = GMm/r², so mv² = GMm/r. KE = ½mv² = GMm/(2r). Total energy E = KE + PE = GMm/(2r) + (-GMm/r) = -GMm/(2r).
Three identical masses $m$ are arranged at the vertices of an equilateral triangle with side length $a$ . What is the magnitude of the gravitational field at the center of the triangle due to all three masses?
Answer: $0$ The center of an equilateral triangle is equidistant from all three vertices. Each mass contributes a gravitational field of magnitude $\frac{Gm}{r^2}$ toward itself, where $r = \frac{a}{\sqrt{3}}$ is the distance from center to vertex. However, by symmetry, the three field vectors are separated by $120°$ and their vector sum is zero. This is a classic result that students often miss by trying to add magnitudes instead of vectors. Options A, B, and C represent attempts to add the field magnitudes rather than the field vectors.
A comet follows an elliptical orbit around the Sun. At its closest approach (perihelion), it is $1.0 \times 10^{11} \text{ m}$ from the Sun and experiences a gravitational field strength of $5.9 \times 10^{-3} \text{ N kg}^{-1}$ . At its farthest point (aphelion), it is $4.0 \times 10^{11} \text{ m}$ from the Sun. What is the ratio of the comet's speeds $\frac{v_{perihelion}}{v_{aphelion}}$ ?
Answer: $4.0$ Using conservation of angular momentum for the elliptical orbit: $L = mr_p v_p = mr_a v_a$ , so $\frac{v_p}{v_a} = \frac{r_a}{r_p} = \frac{4.0 \times 10^{11}}{1.0 \times 10^{11}} = 4.0$ . The gravitational field information is provided as a distractor. Option A uses $\sqrt{4} = 2$ , option C might come from incorrectly using the field strength data, and option D uses $(r_a/r_p)^2 = 16$ .
A spacecraft travels from Earth's surface to the Moon's surface. Given that Earth's mass is $81$ times the Moon's mass and Earth's radius is $3.7$ times the Moon's radius, what is the ratio of the escape speeds $\frac{v_{Earth}}{v_{Moon}}$ ?
Answer: $4.7$ Escape speed is $v_{escape} = \sqrt{\frac{2GM}{R}}$ . Therefore $\frac{v_{Earth}}{v_{Moon}} = \sqrt{\frac{M_{Earth}}{M_{Moon}} \times \frac{R_{Moon}}{R_{Earth}}} = \sqrt{81 \times \frac{1}{3.7}} = \sqrt{21.9} = 4.7$ . Option B might result from using $\sqrt{81/3.7}$ incorrectly, option C uses the mass ratio directly, and option D uses $81/3.7$ without taking the square root.
A satellite in a circular orbit fires its engine, causing it to slow down. Which of the following describes the satellite's subsequent motion, assuming it does not crash?
Answer: It moves into a lower elliptical orbit with the same apogee. Slowing down (retro-firing) reduces the satellite's kinetic and total energy. This causes it to fall into a lower orbit. The point where the engine fired becomes the highest point (apogee) of the new elliptical orbit. The new orbit will have a lower perigee (closest point). Thus, it moves into a lower elliptical orbit with the same apogee as the radius of the original circular orbit.
Two planets, P1 and P2, orbit the same star. The orbital period of P1 is T and its mean orbital radius is R. The mean orbital radius of P2 is 4R. What is the orbital period of P2?
Answer: 8T According to Kepler's Third Law, the square of the orbital period is proportional to the cube of the mean orbital radius (T² ∝ R³). So, (T₂/T₁)² = (R₂/R₁)³. Given R₂ = 4R₁, we have (T₂/T)² = (4R/R)³ = 4³ = 64. Therefore, T₂² = 64T², and T₂ = √64 * T = 8T.
An object is moved from the surface of the Earth (radius R) to an altitude of R above the surface. What is the change in its gravitational potential energy?
Answer: mgR/2 Gravitational potential energy is $E_p$ = -GMm/r. At the surface, r₁ = R, so E_p1 = -GMm/R. At altitude R, the new distance from the center is r₂ = 2R, so E_p2 = -GMm/(2R). The change is Δ $E_p$ = E_p2 - E_p1 = (-GMm/(2R)) - (-GMm/R) = GMm/R - GMm/(2R) = GMm/(2R). Since g = GM/R² on the surface, we can write GM = gR². Substituting this gives Δ $E_p$ = (gR²m)/(2R) = mgR/2.
The total energy of a satellite of mass m in a circular orbit of radius r around a planet of mass M is E = -GMm/(2r). What is the minimum work that must be done by an external agent to move this satellite to an orbit of radius 2r?
Answer: GMm/(4r) Work done is the change in total energy, W = ΔE = E_final - E_initial. The initial energy is E_initial = -GMm/(2r). The final energy in the orbit of radius 2r is E_final = -GMm/(2(2r)) = -GMm/(4r). The work done is W = (-GMm/(4r)) - (-GMm/(2r)) = -GMm/(4r) + 2GMm/(4r) = GMm/(4r).
A spacecraft is at a point in space where the gravitational fields from the Earth and the Moon cancel out. Let $M_E$ be the mass of the Earth, $M_M$ the mass of the Moon, and D the distance between their centers. How far is this point from the center of the Earth?
Answer: D / (1 + √( $M_M$ / $M_E$ )) Let the distance from Earth be x. The distance from the Moon is D-x. The gravitational fields must be equal in magnitude: GM_E/x² = GM_M/(D-x)². This simplifies to $M_E$ /x² = $M_M$ /(D-x)². Rearranging gives (D-x)²/x² = $M_M$ / $M_E$ . Taking the square root: (D-x)/x = √( $M_M$ / $M_E$ ). Then D/x - 1 = √( $M_M$ / $M_E$ ), so D/x = 1 + √( $M_M$ / $M_E$ ). Finally, x = D / (1 + √( $M_M$ / $M_E$ )).
Planet X has mass M and radius R. Planet Y has mass 2M and radius 2R. An object has weight W when on the surface of Planet X. What is the weight of the same object on the surface of Planet Y?
Answer: W/2 Weight is the gravitational force, which is proportional to M/R². For Planet X, W ∝ M/R². For Planet Y, the new weight W' ∝ (2M)/(2R)² = 2M/(4R²) = (1/2) * (M/R²). Therefore, the new weight W' is W/2.
Two spherical masses, m₁ and m₂, are separated by a distance r. The magnitude of the gravitational force between them is F. If the distance is doubled to 2r and the mass m₁ is also doubled to 2m₁, what is the new magnitude of the gravitational force?
Answer: F/2 The original force is F = Gm₁m₂/r². The new force F' = G(2m₁)m₂/(2r)² = G(2m₁m₂)/(4r²) = (2/4) * (Gm₁m₂/r²) = (1/2)F. The new force is F/2.
Two identical satellites orbit a planet at different altitudes. Satellite A has an orbital period of $4.0 \text{ hours}$ and experiences a gravitational field strength of $6.0 \text{ N kg}^{-1}$ . Satellite B has an orbital period of $8.0 \text{ hours}$ . What is the gravitational field strength experienced by satellite B?
Answer: $2.4 \text{ N kg}^{-1}$ From Kepler's third law, $\frac{T_B^2}{T_A^2} = \frac{r_B^3}{r_A^3}$ , so $\frac{64}{16} = \frac{r_B^3}{r_A^3}$ , giving $\frac{r_B}{r_A} = 4^{1/3} = 1.587$ . Since gravitational field strength $g \propto \frac{1}{r^2}$ : $\frac{g_B}{g_A} = \frac{r_A^2}{r_B^2} = \frac{1}{(1.587)^2} = \frac{1}{2.52} = 0.397$ . Therefore $g_B = 6.0 \times 0.397 = 2.4 \text{ N kg}^{-1}$ . Option A uses $\frac{1}{4}$ ratio incorrectly, C uses $\frac{1}{2}$ , and D uses incorrect proportionality.
The gravitational field lines around a planet show the direction of the force on a test mass. Which statement correctly describes the equipotential surfaces associated with this field?
Answer: Equipotential surfaces are spaced further apart where the gravitational field is weaker. Equipotential surfaces are surfaces of constant potential. The gravitational field is stronger where these surfaces are closer together. Conversely, they are spaced further apart where the field is weaker. Field lines are always perpendicular to equipotential surfaces. No work is done moving a mass along an equipotential surface (W = mΔV = 0). Gravitational potential V = -GM/r, which becomes less negative (increases) as r increases.
Two stars of equal mass M form a binary system, orbiting their common centre of mass in a circle of radius R. The distance between the stars is 2R. What is the orbital speed of each star?
Answer: √(GM/(4R)) The gravitational force on one star due to the other provides the centripetal force. The distance between them is d = 2R. The force is F = GM²/(2R)². The centripetal force required for a star of mass M to move in a circle of radius R is $F_c$ = Mv²/R. Setting F = $F_c$ : GM²/(4R²) = Mv²/R. Simplifying by cancelling M and one R gives GM/(4R) = v². Therefore, v = √(GM/(4R)).
A satellite is in a stable circular orbit of radius R around a planet. Its orbital speed is v. If the satellite is moved to a new stable circular orbit of radius 4R, what is its new orbital speed?
Answer: v/2 For a circular orbit, the gravitational force provides the centripetal force: GMm/R² = mv²/R. This simplifies to v² = GM/R, so v = √(GM/R). The speed is inversely proportional to the square root of the radius (v ∝ 1/√R). If the radius R is multiplied by 4, the new speed will be multiplied by 1/√4 = 1/2. The new speed is v/2.

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