Exercises — Gravitation

21 questions✓ Free · step-by-step

7.15 marksNCERT Cl-11 Physics Part I, Ch7 Exercises, Q7.1
Answer the following: (a) You can shield a charge from electrical forces by putting it inside a hollow conductor. Can you shield a body from the gravitational influence of nearby matter by putting it inside a hollow sphere or by some other means? (b) An astronaut inside a small space ship orbiting around the earth cannot detect gravity. If the space station orbiting around the earth has a large size, can he hope to detect gravity? (c) If you compare the gravitational force on the earth due to the sun to that due to the moon, you would find that the Sun's pull is greater than the moon's pull. However, the tidal effect of the moon's pull is greater than the tidal effect of sun. Why?
Hint. For (c), remember that tides come from the DIFFERENCE in pull across the size of an object, not from the pull's overall strength — and that difference falls off faster with distance than the force itself.
Step 1 — (a) Gravitational shielding. No. Electrical shielding works because a conductor has free charges that rearrange to cancel the field inside. Gravity has no negative mass to play an equivalent role, so there is no known way to block or shield gravitational influence — a hollow sphere placed around a body does not stop outside matter from pulling on it.

Step 2 — (b) Detecting gravity in a large station. Yes. A small spacecraft is in free fall, so it and everything inside it fall together and gravity seems to vanish locally. But if the station is very large, the gravitational field varies noticeably from one side of it to the other (it is slightly stronger on the side closer to Earth), and this difference — a tidal effect — becomes detectable even though the station as a whole is still in free fall.

Step 3 — (c) Why the Moon's tides beat the Sun's despite its weaker pull. Tidal effect depends on how much the gravitational pull differs across the width of the object, and this difference falls off as 1/r³ with distance, much faster than the force itself (which falls off as 1/r²). The Moon is far closer to Earth than the Sun is, so even though the Sun's overall pull on Earth is stronger, the Moon's pull varies much more sharply from Earth's near side to its far side, giving it the larger tidal effect.

✦ Answer: (a) No, gravity cannot be shielded (b) yes, tidal (differential) effects become detectable in a large enough station (c) tidal effect depends on the 1/r³ gradient of the field, not the 1/r² force itself, and the Moon's proximity wins out over the Sun's greater mass on that measure.
Where students slip. In (c), assuming tidal strength should simply track which body pulls harder overall — tides are specifically about how unevenly a body pulls across an object's width, a distinct quantity (the field's spatial gradient) from the total force, and the two don't have to rank the same way.
7.24 marksNCERT Cl-11 Physics Part I, Ch7 Exercises, Q7.2
Choose the correct alternative: (a) Acceleration due to gravity increases/decreases with increasing altitude. (b) Acceleration due to gravity increases/decreases with increasing depth (assume the earth to be a sphere of uniform density). (c) Acceleration due to gravity is independent of mass of the earth/mass of the body. (d) The formula -GMm(1/r2 - 1/r1) is more/less accurate than the formula mg(r2-r1) for the difference of potential energy between two points r2 and r1 distance away from the centre of the earth.
Hint. For (d), remember that mgh assumes g stays constant between the two points — check whether that's actually a safe assumption here.
Step 1 — (a) With altitude. g = GM/r², and r increases with altitude, so g decreases.

Step 2 — (b) With depth. Inside a uniform-density Earth, g is proportional to r itself (the distance from the centre), so g decreases as depth increases (r decreases toward the centre).

Step 3 — (c) Dependence on mass. g = GM_earth/r² does not involve the mass of the falling body at all, so it is independent of the body's mass — but it does depend on the Earth's own mass.

Step 4 — (d) Comparing the two potential energy formulas. The exact formula −GMm(1/r2 − 1/r1) correctly accounts for how g itself changes between r1 and r2, while mg(r2−r1) assumes g stays constant over that whole interval — an approximation that is only good for small height differences. The exact formula is more accurate.

✦ Answer: (a) decreases (b) decreases (c) independent of mass of the body (d) more accurate.
Where students slip. In (c), answering 'independent of mass of the earth' — g clearly depends on the Earth's own mass (it's right there in the formula GM/r²); it's only independent of the mass of whatever body happens to be falling.
7.33 marksNCERT Cl-11 Physics Part I, Ch7 Exercises, Q7.3
Suppose there existed a planet that went around the Sun twice as fast as the earth. What would be its orbital size as compared to that of the earth?
Hint. Apply Kepler's third law directly as a ratio between the hypothetical planet and Earth, so that the constant of proportionality cancels out.
Step 1 — Set up the period ratio. Going around twice as fast means half the orbital period: T_planet = T_earth/2.

Step 2 — Apply Kepler's third law as a ratio. (T_planet/T_earth)² = (R_planet/R_earth)³, so (1/2)² = (R_planet/R_earth)³.

Step 3 — Solve for the radius ratio. 1/4 = (R_planet/R_earth)³, so R_planet/R_earth = (1/4)^(1/3) ≈ 0.63.

✦ Answer: The planet's orbital radius would be about 0.63 times that of Earth's — a smaller, faster orbit.
Where students slip. Assuming twice the speed simply means half the orbital radius, treating T and R as directly proportional — Kepler's third law relates T² to R³, not T to R, so halving the period only shrinks the radius by a factor of (1/2)^(2/3), not (1/2) itself.
7.45 marksNCERT Cl-11 Physics Part I, Ch7 Exercises, Q7.4
Io, one of the satellites of Jupiter, has an orbital period of 1.769 days and the radius of the orbit is 4.22x10^8 m. Show that the mass of Jupiter is about one-thousandth that of the sun.
Hint. Use Kepler's third law in its full form, M = 4(pi)^2 R^3 / (G T^2), to find Jupiter's mass directly from Io's orbit, then compare to the Sun's well-known mass.
Step 1 — Convert the period to seconds. T = 1.769 days × 86,400 s/day ≈ 152,842 s.

Step 2 — Apply Kepler's third law to find Jupiter's mass. M = 4π²R³/(GT²). With R = 4.22×10⁸ m: R³ ≈ 7.515×10²⁵ m³, and T² ≈ 2.336×10¹⁰ s².

Step 3 — Compute. M ≈ (4π² × 7.515×10²⁵) / (6.67×10⁻¹¹ × 2.336×10¹⁰) ≈ 2.967×10²⁷ / 1.558 ≈ 1.90×10²⁷ kg.

Step 4 — Compare with the Sun's mass. The Sun's mass is about 1.99×10³⁰ kg, so M_Jupiter/M_Sun ≈ 1.90×10²⁷/1.99×10³⁰ ≈ 9.6×10⁻⁴, which is close to 1/1000.

✦ Answer: Jupiter's mass works out to about 1.9×10²⁷ kg — roughly one-thousandth of the Sun's mass, exactly as the question asks to show.
Where students slip. Forgetting to convert Io's period from days to seconds before squaring it — every quantity in Kepler's law here (G, R, T) needs to be in consistent SI units, and leaving T in days would throw the mass off by many orders of magnitude.
7.55 marksNCERT Cl-11 Physics Part I, Ch7 Exercises, Q7.5
Let us assume that our galaxy consists of 2.5x10^11 stars each of one solar mass. How long will a star at a distance of 50,000 ly from the galactic centre take to complete one revolution? Take the diameter of the Milky Way to be 10^5 ly.
Hint. Since the star sits at essentially the galaxy's outer edge, treat the combined mass of all the stars as if concentrated at the centre, and apply the same orbital-period formula used for planets around the Sun.
Step 1 — Total mass of the galaxy. M = 2.5×10¹¹ × 2×10³⁰ kg = 5×10⁴¹ kg (using solar mass ≈ 2×10³⁰ kg).

Step 2 — Convert the orbital radius to metres. r = 50,000 ly × 9.46×10¹⁵ m/ly ≈ 4.73×10²⁰ m.

Step 3 — Apply T = 2π√(r³/GM). r³ ≈ 1.058×10⁶² m³. GM ≈ 6.67×10⁻¹¹ × 5×10⁴¹ ≈ 3.335×10³¹. r³/GM ≈ 3.17×10³⁰.

Step 4 — Compute T. √(3.17×10³⁰) ≈ 1.78×10¹⁵. T = 2π × 1.78×10¹⁵ ≈ 1.12×10¹⁶ s.

Step 5 — Convert to years. T ≈ 1.12×10¹⁶ / 3.156×10⁷ ≈ 3.5×10⁸ years, since dividing by the number of seconds in a year gives the period directly in years.

✦ Answer: The star takes about 3.5×10⁸ years (350 million years) to complete one revolution.
Where students slip. Treating the galaxy's diameter (10⁵ ly) as the orbital radius instead of using the given 50,000 ly directly — the diameter is only given to confirm the star sits near the galaxy's edge; the actual orbital radius to use is the 50,000 ly stated explicitly.
7.64 marksNCERT Cl-11 Physics Part I, Ch7 Exercises, Q7.6
Choose the correct alternative: (a) If the zero of potential energy is at infinity, the total energy of an orbiting satellite is negative of its kinetic/potential energy. (b) The energy required to launch an orbiting satellite out of earth's gravitational influence is more/less than the energy required to project a stationary object at the same height (as the satellite) out of earth's influence.
Hint. For (a), recall the standard relation between total orbital energy and kinetic energy for a circular orbit (E = -KE, a form of the virial theorem). For (b), consider what head start an orbiting satellite already has.
Step 1 — (a) Total energy vs kinetic/potential energy. For a circular orbit with the zero of potential energy at infinity, the standard result is E_total = −KE (equivalently, E_total = PE/2) — so the total energy is the negative of the kinetic energy, not the potential energy.

Step 2 — (b) Comparing launch energies. An orbiting satellite already has orbital kinetic energy working in its favour, so it needs less additional energy to reach escape than a stationary object at the same height, which starts with none of that head start.

✦ Answer: (a) negative of its kinetic energy (b) less energy is required for the orbiting satellite than for the stationary object.
Where students slip. Picking 'potential energy' in (a) — the well-known relation for a bound circular orbit is E_total = −KE = PE/2, so the total energy equals the negative of the KINETIC energy specifically, not the potential energy (though it is exactly half the potential energy in magnitude).
7.74 marksNCERT Cl-11 Physics Part I, Ch7 Exercises, Q7.7
Does the escape speed of a body from the earth depend on (a) the mass of the body, (b) the location from where it is projected, (c) the direction of projection, (d) the height of the location from where the body is launched?
Hint. Write out the escape speed formula explicitly and check, term by term, which of the four listed quantities actually appears in it.
Step 1 — Write the escape speed formula. v_esc = √(2GM/R), where M is Earth's mass and R is the distance from Earth's centre at the point of launch.

Step 2 — (a) Mass of the body. The body's own mass doesn't appear in the formula at all — no dependence.

Step 3 — (b) and (d) Location and height. At a fixed distance from Earth's centre, escape speed is the same everywhere (no dependence on which point on that sphere). But height changes R directly, and since v_esc depends on R, escape speed does depend on height (it decreases as height increases).

Step 4 — (c) Direction of projection. The formula has no directional dependence — the same minimum speed suffices regardless of which way the body is fired, as long as its path doesn't intersect the Earth itself.

✦ Answer: (a) No (b) No, for locations at the same height (c) No (d) Yes — escape speed decreases with increasing height, since it depends on R.
Where students slip. Conflating (b) and (d) into a single answer — 'location' at a fixed height doesn't matter, but height itself very much does, since it directly changes R in the escape-speed formula; these are two genuinely different questions.
7.85 marksNCERT Cl-11 Physics Part I, Ch7 Exercises, Q7.8
A comet orbits the sun in a highly elliptical orbit. Does the comet have a constant (a) linear speed, (b) angular speed, (c) angular momentum, (d) kinetic energy, (e) potential energy, (f) total energy throughout its orbit? Neglect any mass loss of the comet when it comes very close to the Sun.
Hint. Gravity is a central, conservative force with no external torque about the Sun — sort each quantity by whether it follows directly from that fact or depends on the comet's changing distance from the Sun.
Step 1 — What stays constant, and why. Angular momentum (c) is conserved because gravity, acting along the line to the Sun, exerts no torque about the Sun. Total energy (f) is conserved because gravity is a conservative force and no other forces do work on the comet.

Step 2 — What varies, and why. By Kepler's second law (a direct consequence of constant angular momentum), the comet moves faster when closer to the Sun and slower when farther away — so linear speed (a) and angular speed (b) both vary. Since speed varies, kinetic energy (d) varies too. Since distance from the Sun varies along an ellipse, potential energy (e) varies as well.

✦ Answer: Constant: (c) angular momentum and (f) total energy. Not constant: (a) linear speed, (b) angular speed, (d) kinetic energy, (e) potential energy.
Where students slip. Assuming kinetic and potential energy must each stay constant just because total energy does — total energy staying fixed is exactly why the two constantly trade off against each other as the comet's distance from the Sun changes, not evidence that either one alone is constant.
7.93 marksNCERT Cl-11 Physics Part I, Ch7 Exercises, Q7.9
Which of the following symptoms is likely to afflict an astronaut in space (a) swollen feet, (b) swollen face, (c) headache, (d) orientational problem.
Hint. In microgravity, bodily fluids no longer get pulled downward and pooled in the lower body the way they are on Earth — think about where that fluid ends up instead.
Step 1 — What happens to body fluids in microgravity. Without gravity pulling fluids toward the feet, blood and other fluids shift upward toward the head and upper body instead of pooling in the legs and feet.

Step 2 — Consequences of this fluid shift. This produces facial puffiness (swollen face) and increased pressure in the head (headache) — both well documented in real astronauts — while the feet, receiving less fluid than usual, do not swell.

Step 3 — Orientation. Since there's no consistent gravitational 'down' to orient by anymore, astronauts commonly experience spatial disorientation, especially early in a mission.

✦ Answer: (b), (c), and (d) are likely; (a) swollen feet is not — if anything, the feet tend to look thinner in microgravity as fluid shifts away from them.
Where students slip. Assuming swollen feet by analogy with long flights or standing on Earth, where gravity pools fluid downward — in microgravity there is no 'downward' pull at all, so the fluid shift actually runs the opposite way, toward the head.
7.104 marksNCERT Cl-11 Physics Part I, Ch7 Exercises, Q7.10
The gravitational intensity at the centre of a hemispherical shell of uniform mass density has the direction indicated by the arrow (see Fig 7.11): (i) a, (ii) b, (iii) c, (iv) 0.
Hint. The shell is symmetric about the vertical axis through its centre and pole — any sideways pull from one part of the bowl is cancelled by the opposite side, leaving only a pull along that axis, toward the bulk of the material.
Step 1 — Use the rotational symmetry of the bowl. The hemispherical shell is symmetric under rotation about the vertical axis passing through its centre C and the lowest point of the bowl. Any sideways (horizontal) pull from one part of the shell is exactly cancelled by the diametrically opposite part.

Step 2 — What direction survives. Only the component along the vertical axis of symmetry can survive this cancellation. Since all of the shell's mass lies below the flat rim plane (curving downward into the bowl), the net pull at C must be directed downward, into the bowl.

✦ Answer: (iii) c — the gravitational intensity at the centre points straight down, into the bowl, along arrow c.
Where students slip. Picking arrow a (horizontal) — by the rotational symmetry of the bowl about the vertical axis, any horizontal pull is guaranteed to cancel out exactly; only a vertical, axial component can possibly survive.
7.115 marksNCERT Cl-11 Physics Part I, Ch7 Exercises, Q7.11
For the above problem, the direction of the gravitational intensity at an arbitrary point P (on the rim of the hemispherical shell) is indicated by the arrow: (i) d, (ii) e, (iii) f, (iv) g.
Hint. Imagine completing the hemisphere into a full spherical shell by adding its mirror image on the other side of the rim plane — the known field of a full shell then lets you extract the piece contributed by just the original half.
Step 1 — Complete the shell by symmetry. Add a mirror-image hemisphere (reflected across the flat rim plane) to the original, forming a complete, uniform spherical shell of the same radius, with P sitting exactly on the rim — which is also the equatorial plane of this completed sphere.

Step 2 — Use the known field of a full shell at its own surface. For a uniform spherical shell, the field just inside is zero and just outside is GM/R² directed toward the centre; at the surface itself the physically meaningful value is the average of the two, GM/(2R²), directed toward the centre.

Step 3 — Split this into the two hemispheres' contributions. Because P lies exactly in the mirror (rim) plane, reflecting the setup leaves P fixed, keeps the horizontal (toward-centre) component of each hemisphere's field the same, but flips the sign of any vertical component. Adding the original and mirrored contributions together must reproduce the full shell's purely horizontal, toward-centre field — which is only possible if the two hemispheres' horizontal components are equal (splitting that total field evenly) while their vertical components exactly cancel.

Step 4 — Read off the direction. This leaves the original hemisphere's own field at P purely horizontal, pointing from P toward the centre C.

✦ Answer: (iii) f — the gravitational intensity at P points horizontally from P toward the centre C.
Where students slip. Assuming the field at P must point straight down (arrow e), by analogy with the field at the centre C — P sits at a very different, less symmetric location (the rim, not the axis), and the completed-sphere argument shows its field is purely horizontal toward C, not vertical.
7.124 marksNCERT Cl-11 Physics Part I, Ch7 Exercises, Q7.12
A rocket is fired from the earth towards the sun. At what distance from the earth's centre is the gravitational force on the rocket zero? Mass of the sun = 2x10^30 kg, mass of the earth = 6x10^24 kg. Neglect the effect of other planets etc. (orbital radius = 1.5x10^11 m).
Hint. Set the two pulls equal at the unknown point, and let x be its distance from Earth's centre so that the remaining distance to the Sun is the total Earth-Sun separation minus x.
Step 1 — Set up the balance condition. At the null point, GM_earth/x² = GM_sun/(1.5×10¹¹ − x)², so M_earth/x² = M_sun/(1.5×10¹¹−x)².

Step 2 — Take the square root of the mass ratio. (1.5×10¹¹−x)/x = √(M_sun/M_earth) = √(2×10³⁰/6×10²⁴) = √(3.33×10⁵) ≈ 577.4.

Step 3 — Solve for x. 1.5×10¹¹ − x = 577.4x → 1.5×10¹¹ = 578.4x → x ≈ 2.59×10⁸ m.

✦ Answer: The gravitational force is zero at about 2.6×10⁸ m from Earth's centre, along the line toward the Sun.
Where students slip. Assuming the null point sits roughly halfway between Earth and the Sun — since the Sun is vastly more massive, its pull only becomes comparable to Earth's very close to Earth itself, not anywhere near the midpoint of the full 1.5×10¹¹ m separation.
7.134 marksNCERT Cl-11 Physics Part I, Ch7 Exercises, Q7.13
How will you 'weigh the sun', that is estimate its mass? The mean orbital radius of the earth around the sun is 1.5x10^8 km.
Hint. Kepler's third law, rearranged, gives the central mass directly from any orbiting body's period and orbital radius — Earth's own year and orbital radius are exactly what's needed.
Step 1 — Rearrange Kepler's third law for mass. Since T² = 4π²R³/(GM), the Sun's mass is M = 4π²R³/(GT²).

Step 2 — Substitute Earth's orbital data. R = 1.5×10¹¹ m, T = 1 year ≈ 3.156×10⁷ s. R³ ≈ 3.375×10³³ m³, T² ≈ 9.96×10¹⁴ s².

Step 3 — Compute. M = (4π² × 3.375×10³³)/(6.67×10⁻¹¹ × 9.96×10¹⁴) ≈ 1.332×10³⁵/6.64×10⁴ ≈ 2.0×10³⁰ kg.

✦ Answer: Measuring Earth's orbital radius and period and applying Kepler's third law gives the Sun's mass as about 2.0×10³⁰ kg.
Where students slip. Trying to weigh the Sun using a force-balance argument without a known orbital period — the method that actually works uses Kepler's third law, which needs both the orbital radius AND the orbital period together; radius alone isn't enough to isolate the mass.
7.143 marksNCERT Cl-11 Physics Part I, Ch7 Exercises, Q7.14
A saturn year is 29.5 times the earth year. How far is the saturn from the sun if the earth is 1.50x10^8 km away from the sun?
Hint. Apply Kepler's third law as a direct ratio between Saturn and Earth, so neither G nor the Sun's mass needs to be known explicitly.
Step 1 — Set up the ratio form of Kepler's third law. (T_Saturn/T_Earth)² = (R_Saturn/R_Earth)³.

Step 2 — Substitute the period ratio. (29.5)² = (R_Saturn/R_Earth)³, so 870.25 = (R_Saturn/R_Earth)³.

Step 3 — Solve for the radius ratio. R_Saturn/R_Earth = 870.25^(1/3) ≈ 9.55.

Step 4 — Compute Saturn's distance. R_Saturn ≈ 9.55 × 1.50×10⁸ km ≈ 1.43×10⁹ km.

✦ Answer: Saturn is about 1.43×10⁹ km from the Sun.
Where students slip. Scaling the distance by 29.5 directly (matching the period ratio) — Kepler's third law relates T² to R³, so the distance ratio is the CUBE ROOT of the squared period ratio, not the period ratio itself.
7.153 marksNCERT Cl-11 Physics Part I, Ch7 Exercises, Q7.15
A body weighs 63 N on the surface of the earth. What is the gravitational force on it due to the earth at a height equal to half the radius of the earth?
Hint. At height R/2, the body's distance from Earth's centre becomes 1.5R — apply the inverse-square scaling directly using this ratio.
Step 1 — Find the new distance from Earth's centre. At height R/2 above the surface, distance from the centre = R + R/2 = 3R/2.

Step 2 — Apply the inverse-square law as a ratio. F_new/F_surface = (R/(3R/2))² = (2/3)² = 4/9.

Step 3 — Compute the new force. F_new = (4/9) × 63 = 28 N, since this fraction is exactly what the squared distance ratio from Step 2 gives.

✦ Answer: The gravitational force at that height is 28 N.
Where students slip. Using R/2 itself as the new distance from the centre instead of R + R/2 — height above the surface adds to the Earth's radius rather than replacing it as the distance from the centre.
7.163 marksNCERT Cl-11 Physics Part I, Ch7 Exercises, Q7.16
Assuming the earth to be a sphere of uniform mass density, how much would a body weigh half way down to the centre of the earth if it weighed 250 N on the surface?
Hint. Inside a uniform-density sphere, gravity's strength scales linearly with distance from the centre, not by the inverse-square law used outside — apply that linear scaling directly.
Step 1 — Recall how g behaves inside a uniform sphere. For r < R, g(r) = g_surface × (r/R) — a linear relationship, unlike the inverse-square law that applies outside the sphere.

Step 2 — Apply it at r = R/2 (halfway to the centre). g(R/2) = g_surface × (1/2).

Step 3 — Compute the new weight. Weight = 250 × (1/2) = 125 N, since g itself has simply halved at that depth.

✦ Answer: The body would weigh 125 N halfway down to the centre.
Where students slip. Applying the inverse-square law (as if still outside the Earth) to get 250 × 4 = 1000 N or similar — the inverse-square law only holds outside a spherically symmetric mass; inside a uniform sphere, gravity instead falls off linearly toward the centre, reaching zero exactly at the centre.
7.175 marksNCERT Cl-11 Physics Part I, Ch7 Exercises, Q7.17
A rocket is fired vertically with a speed of 5 km/s from the earth's surface. How far from the earth does the rocket go before returning to the earth? Mass of the earth = 6.0x10^24 kg; mean radius of the earth = 6.4x10^6 m; G = 6.67x10^-11 N m^2 kg^-2.
Hint. Use energy conservation between the launch point and the highest point (where the rocket is momentarily at rest), rather than the constant-g kinematic equations, since g changes noticeably over such a large altitude range.
Step 1 — Set up energy conservation from launch to maximum distance. ½v² − GM/R = −GM/r_max (kinetic energy is zero at the highest point, r_max being the distance from Earth's centre there).

Step 2 — Rearrange for 1/r_max. 1/r_max = 1/R − v²/(2GM).

Step 3 — Compute each term. GM = 6.67×10⁻¹¹ × 6.0×10²⁴ ≈ 4.0×10¹⁴. v²/(2GM) = (5000)²/(2×4.0×10¹⁴) ≈ 3.12×10⁻⁸. 1/R = 1/(6.4×10⁶) ≈ 1.5625×10⁻⁷.

Step 4 — Solve for r_max. 1/r_max ≈ 1.5625×10⁻⁷ − 3.12×10⁻⁸ ≈ 1.25×10⁻⁷, so r_max ≈ 8.0×10⁶ m.

Step 5 — Convert to height above the surface. Height = r_max − R = 8.0×10⁶ − 6.4×10⁶ = 1.6×10⁶ m.

✦ Answer: The rocket rises to about 1.6×10⁶ m (1600 km) above the Earth's surface before falling back.
Where students slip. Using the constant-acceleration formula v² = u² − 2gh with g = 9.8 m/s² throughout — g drops noticeably over a 1600 km rise, so only the full energy-conservation approach (which properly accounts for the changing gravitational potential) gives the correct answer here.
7.184 marksNCERT Cl-11 Physics Part I, Ch7 Exercises, Q7.18
The escape speed of a projectile on the earth's surface is 11.2 km/s. A body is projected out with thrice this speed. What is the speed of the body far away from the earth? Ignore the presence of the sun and other planets.
Hint. Recall that escape speed is defined by GM/R = half the escape speed squared — substitute this relation into the energy-conservation equation to avoid needing G, M, or R explicitly.
Step 1 — Set up energy conservation from the surface to far away. ½v_launch² − GM/R = ½v_final² (potential energy → 0 far away).

Step 2 — Use the definition of escape speed. Since ½v_esc² = GM/R by definition, substitute: ½v_launch² − ½v_esc² = ½v_final², so v_final² = v_launch² − v_esc².

Step 3 — Substitute v_launch = 3v_esc. v_final² = (3v_esc)² − v_esc² = 9v_esc² − v_esc² = 8v_esc².

Step 4 — Solve for v_final. v_final = v_esc√8 = 11.2 × 2.828 ≈ 31.7 km/s.

✦ Answer: The body's speed far from Earth is about 31.7 km/s.
Where students slip. Computing v_final as simply 3 times the escape speed minus something, or assuming it scales linearly with launch speed — the relationship comes from squares (kinetic energies), so the final speed is √8 ≈ 2.83 times the escape speed, not related to the factor of 3 in any simpler way.
7.194 marksNCERT Cl-11 Physics Part I, Ch7 Exercises, Q7.19
A satellite orbits the earth at a height of 400 km above the surface. How much energy must be expended to rocket the satellite out of the earth's gravitational influence? Mass of the satellite = 200 kg; mass of the earth = 6.0x10^24 kg; radius of the earth = 6.4x10^6 m; G = 6.67x10^-11 N m^2 kg^-2.
Hint. A satellite in circular orbit has total energy -GMm/(2r) — the energy needed to escape is simply whatever must be added to bring that total up to zero.
Step 1 — Find the orbital radius. r = R + height = 6.4×10⁶ + 0.4×10⁶ = 6.8×10⁶ m.

Step 2 — Recall the total energy of a circular orbit. E_orbit = −GMm/(2r).

Step 3 — Find the energy needed to escape. Escaping to infinity means reaching E = 0, so the energy that must be supplied is ΔE = 0 − E_orbit = GMm/(2r).

Step 4 — Compute. GMm = 6.67×10⁻¹¹ × 6.0×10²⁴ × 200 ≈ 8.0×10¹⁶. Dividing by 2r = 1.36×10⁷: ΔE ≈ 8.0×10¹⁶/1.36×10⁷ ≈ 5.9×10⁹ J.

✦ Answer: About 5.9×10⁹ J (5.9 GJ) must be expended to send the satellite out of Earth's gravitational influence.
Where students slip. Using the full GMm/r (without the factor of 2) as the orbital energy — a satellite in a BOUND circular orbit has total energy exactly half the magnitude of GMm/r, since half of that magnitude is already 'spent' as kinetic energy keeping it in orbit.
7.205 marksNCERT Cl-11 Physics Part I, Ch7 Exercises, Q7.20
Two stars each of one solar mass (= 2x10^30 kg) are approaching each other for a head on collision. When they are a distance 10^9 km, their speeds are negligible. What is the speed with which they collide? The radius of each star is 10^4 km. Assume the stars to remain undistorted until they collide. (Use the known value of G).
Hint. By symmetry, equal masses starting from rest and pulled only by their mutual gravity always move with equal speeds toward each other — use energy conservation between the initial separation and the moment their surfaces touch.
Step 1 — Identify the initial and final separations. Initial: d_i = 10⁹ km = 10¹² m (speeds negligible, so KE ≈ 0). Final (surfaces touching): d_f = 2 × 10⁴ km = 2×10⁷ m (sum of the two radii).

Step 2 — Set up energy conservation. By symmetry both stars always have equal speed v. −Gm²/d_i = 2(½mv²) − Gm²/d_f.

Step 3 — Solve for v². v² = Gm(1/d_f − 1/d_i). Since 1/d_i (≈10⁻¹²) is negligible next to 1/d_f (≈5×10⁻⁸), v² ≈ Gm/d_f.

Step 4 — Compute. v² ≈ 6.67×10⁻¹¹ × 2×10³⁰ × 5×10⁻⁸ ≈ 6.67×10¹². v ≈ √(6.67×10¹²) ≈ 2.58×10⁶ m/s.

✦ Answer: Each star collides at a speed of about 2.58×10⁶ m/s.
Where students slip. Forgetting the factor of 2 for kinetic energy (both stars are moving, not just one) — by the symmetry of two equal masses with no external forces, both share the released potential energy equally as kinetic energy, not just one of them.
7.215 marksNCERT Cl-11 Physics Part I, Ch7 Exercises, Q7.21
Two heavy spheres each of mass 100 kg and radius 0.10 m are placed 1.0 m apart on a horizontal table. What is the gravitational force and potential at the mid point of the line joining the centres of the spheres? Is an object placed at that point in equilibrium? If so, is the equilibrium stable or unstable?
Hint. Force is a vector and can cancel by symmetry; potential is a scalar and always adds — treat these two parts of the question separately, and then think about what happens to the force balance under a small nudge away from the midpoint.
Step 1 — Force at the midpoint. The midpoint is equidistant (0.5 m) from both equal masses, and the two pulls point in exactly opposite directions along the line joining the centres — they cancel exactly, giving zero net force.

Step 2 — Potential at the midpoint. Potential is a scalar, so the two contributions add rather than cancel: V = −Gm/r − Gm/r = −2Gm/r = −2(6.67×10⁻¹¹)(100)/0.5 ≈ −2.67×10⁻⁸ J/kg.

Step 3 — Is it in equilibrium? Yes — since the net force there is exactly zero, an object placed at the midpoint is in equilibrium.

Step 4 — Stable or unstable? Displace the object slightly toward one sphere: that sphere's pull grows stronger (closer) while the other's grows weaker (farther), so the net force now pulls the object further away from the midpoint, not back toward it — the equilibrium is unstable.

✦ Answer: Force = 0; potential ≈ −2.67×10⁻⁸ J/kg. The object is in equilibrium, but the equilibrium is unstable, since any small displacement along the line joining the centres grows rather than being restored.
Where students slip. Concluding the equilibrium must be stable just because the net force is zero there — zero force only means the point is an equilibrium; whether it's stable or unstable depends on what happens to the force under a small displacement, which here pulls the object further away rather than back.

Solutions written by the tuition.in editorial team and checked against the NCERT Class 11 Physics Part I textbook, Reprint 2026-27 (keph107.pdf) — one end-of-chapter Exercises set (21 questions, 7.1-7.21); one page (page 15 of the PDF) failed direct text extraction due to a font-encoding issue and was instead rendered as an image and read visually in full, including Fig 7.11's hemispherical-shell arrow diagram used for Q7.10/7.11, verified via a 600dpi zoom. Questions are referenced from the NCERT textbook for identification.

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