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The Vault
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The Vault
CBSE Class 11 Physics · 10 questions · 24 marks
The same force that makes an apple fall keeps the Moon in orbit, and that single realisation is the heart of this chapter. You will move from Kepler's descriptive laws of planetary motion to Newton's universal law that explains them, then use gravitational potential energy to work out escape velocity and the physics of satellites. Careful attention to signs in potential energy is what separates a correct answer from a confusing one here.
The escape velocity from the surface of a planet does NOT depend on:
Answer
The mass of the escaping body — from v_e = √(2GM/R), only the planet's mass M, its radius R and the constant G appear. The escaping body's own mass cancels because it multiplies both its kinetic energy and its gravitational potential energy equally. A pebble and a spacecraft therefore need the same escape speed, though of course they need very different amounts of energy.
A planet's orbital radius around a star is increased by a factor of 4. Its period of revolution becomes:
Answer
8 times the original — by Kepler's third law T² ∝ r³, so T ∝ r^(3/2). If r becomes 4r, then T becomes 4^(3/2) = 8 times the original. Assuming T ∝ r gives the wrong answer of 4.
The value of the acceleration due to gravity at the centre of the Earth, assuming uniform density, is:
Answer
Zero — inside a uniform sphere, g_d = g(1 − d/R). At the centre d = R, so g_d = 0. Physically, only the mass in the sphere below you pulls on you, and at the centre there is no such mass; the surrounding shell exerts zero net force on anything inside it. The answer 'infinite' comes from wrongly applying the inverse-square formula for external points to the interior.
The gravitational potential energy of a satellite of mass m in a circular orbit of radius r around the Earth of mass M is −GMm/r. Its total energy is:
Answer
−GMm/2r — the orbital condition GMm/r² = mv²/r gives kinetic energy K = ½mv² = GMm/2r. Adding the potential energy: E = GMm/2r − GMm/r = −GMm/2r. The total energy is negative, which is the signature of a bound orbit, and its magnitude equals the kinetic energy.
Assertion (A): An astronaut inside an orbiting space station floats and feels weightless. Reason (R): There is no gravitational force acting on the astronaut at that height.
Answer
A is true but R is false — the astronaut does float, but not because gravity is absent. At a typical station height of about 400 km, g is still roughly 89% of its surface value. The astronaut feels weightless because the station and everything in it are in free fall together, accelerating towards the Earth at the same rate; with no normal reaction from the floor there is no sensation of weight. Weightlessness in orbit is the absence of a contact force, not the absence of gravity.
Calculate the escape velocity from the surface of a planet whose radius is 3.2×10⁶ m and whose surface gravity is 5 m/s².
Answer
Using v_e = √(2gR): v_e = √(2 × 5 × 3.2×10⁶) = √(3.2×10⁷) v_e = √(32×10⁶) ≈ 5.66×10³ m/s. So the escape velocity is about 5.7 km/s, roughly half the Earth's value, as expected for a smaller and less massive planet.
At what height above the Earth's surface does the acceleration due to gravity fall to one quarter of its surface value? Take the Earth's radius as 6400 km.
Answer
At height h, g_h = GM/(R + h)² and at the surface g = GM/R². Requiring g_h = g/4: GM/(R + h)² = GM/(4R²) (R + h)² = 4R² R + h = 2R, so h = R = 6400 km. The distance from the centre has doubled, and by the inverse-square law doubling the distance quarters the field. Note this exact relation must be used rather than the approximation g_h = g(1 − 2h/R), which is valid only for h ≪ R.
Derive an expression for the escape velocity of a body from the surface of a planet of mass M and radius R using energy conservation. Hence calculate the escape velocity from a moon of mass 7.0×10²² kg and radius 1.7×10⁶ m (G = 6.67×10⁻¹¹ N·m²/kg²).
Answer
Derivation: consider a body of mass m projected vertically upward from the surface of a planet of mass M and radius R with speed v. Escaping means reaching an infinite distance, where by convention the gravitational potential energy is zero, and just barely arriving there, so the final kinetic energy is also zero. Initial total energy at the surface: E_i = ½mv² + (−GMm/R) = ½mv² − GMm/R. Final total energy at infinity: E_f = 0 + 0 = 0. Only gravity, a conservative force, acts, so total mechanical energy is conserved: ½mv² − GMm/R = 0 ½mv² = GMm/R v² = 2GM/R v_e = √(2GM/R). The mass m of the projected body cancels, so escape velocity is a property of the planet alone. Writing GM = gR², this also becomes v_e = √(2gR). Calculation for the given moon: v_e = √(2 × 6.67×10⁻¹¹ × 7.0×10²²/1.7×10⁶) Numerator: 2 × 6.67×10⁻¹¹ × 7.0×10²² = 9.338×10¹². Dividing: 9.338×10¹²/1.7×10⁶ = 5.49×10⁶. v_e = √(5.49×10⁶) ≈ 2.34×10³ m/s, that is about 2.3 km/s. Because this is low, light gas molecules can easily exceed it thermally, which is why such a body retains almost no atmosphere.
(a) State Kepler's three laws of planetary motion. (b) Show that Kepler's second law follows from the conservation of angular momentum. (c) A satellite orbits a planet of mass 6.0×10²⁴ kg in a circular orbit of radius 8.0×10⁶ m. Find its orbital speed (G = 6.67×10⁻¹¹ N·m²/kg²).
Answer
(a) First law (law of orbits): each planet moves along an ellipse with the Sun situated at one of the two foci. Second law (law of areas): the line joining a planet to the Sun sweeps out equal areas in equal intervals of time, so a planet moves fastest at perihelion and slowest at aphelion. Third law (law of periods): the square of the orbital period is proportional to the cube of the semi-major axis, T² ∝ a³. (b) Gravity acts along the line joining the planet to the Sun, so it is a central force and its torque about the Sun is τ = r × F = 0 because r and F are antiparallel. With zero torque, the planet's angular momentum L about the Sun is constant. In a small time dt the planet moves by dr, and the area swept by the radius vector is the area of the thin triangle: dA = ½|r × dr|. Dividing by dt: dA/dt = ½|r × (dr/dt)| = ½|r × v| = |r × mv|/2m = L/2m. Since L and m are both constant, dA/dt is constant. Equal areas are therefore swept in equal times, which is exactly Kepler's second law. (c) For a circular orbit the gravitational force supplies the centripetal force: GMm/r² = mv²/r, so v = √(GM/r). v = √(6.67×10⁻¹¹ × 6.0×10²⁴/8.0×10⁶) Numerator: 6.67×10⁻¹¹ × 6.0×10²⁴ = 4.002×10¹⁴. Dividing: 4.002×10¹⁴/8.0×10⁶ = 5.0×10⁷. v = √(5.0×10⁷) ≈ 7.07×10³ m/s, about 7.1 km/s.
Read the following and answer the questions that follow: A communications company plans a geostationary satellite that must stay fixed above one point on the equator, and also a low Earth observation satellite at a height of 600 km. Take the Earth's mass as 6.0×10²⁴ kg, its radius as 6400 km, g = 9.8 m/s² at the surface and G = 6.67×10⁻¹¹ N·m²/kg². (a) State the two conditions the geostationary orbit must satisfy besides having the correct radius. (b) Calculate the orbital speed of the observation satellite at 600 km. (c) Explain why the total energy of any bound satellite is negative and what it would mean for that energy to become zero. (d) A small thruster fires and slightly reduces the observation satellite's speed. Describe qualitatively what happens to its orbit.
Answer
(a) The orbit must lie exactly in the plane of the Earth's equator, and the satellite must revolve in the same sense as the Earth's rotation (west to east) with a period of exactly one sidereal day, about 24 hours. If the orbit were tilted, the satellite would trace a figure-of-eight in the sky rather than staying put; if it revolved the other way, it would race across the sky twice a day. (b) Orbital radius r = R + h = 6400 km + 600 km = 7000 km = 7.0×10⁶ m. v_o = √(GM/r) = √(6.67×10⁻¹¹ × 6.0×10²⁴/7.0×10⁶). Numerator: 4.002×10¹⁴. Dividing by 7.0×10⁶ gives 5.717×10⁷. v_o = √(5.717×10⁷) ≈ 7.56×10³ m/s, about 7.6 km/s. (c) The total energy is E = K + U = GMm/2r − GMm/r = −GMm/2r. The magnitude of the negative potential energy always exceeds the positive kinetic energy for a closed orbit, so E < 0. This negative sign is the mathematical statement that the satellite is bound: work must be supplied from outside to free it. If E were raised to exactly zero, the satellite would have just enough energy to reach infinity with zero speed — it would be moving at escape velocity and would leave on a parabolic path, never to return. (d) A slight loss of speed means the gravitational pull is momentarily more than the reduced centripetal requirement, so the satellite falls inward. It does not simply drop; it settles into a smaller elliptical orbit whose apogee is near the firing point and whose perigee is lower. Its total energy has become more negative, and by Kepler's third law the smaller orbit has a shorter period, so counterintuitively the satellite ends up moving faster on average than before. If the perigee dips into the upper atmosphere, drag will bleed away more energy and the orbit will decay.
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