Chapter 4 · In-text numericalsTextbook pages 81–82

In-text Numericals – Solutions ഗുരുത്വാകർഷണം – ചോദ്യങ്ങൾ

Four ❓ problems sit inside the text on pages 81–82: a dead satellite falling to the Earth, a 20 m drop on the Earth and on the Moon, a stone thrown up on the Moon, and the weight of a 10 kg object. Each one is worked in exam layout – Given → Formula → Substitution → Answer with unit. Try each one in your notebook first, then step through the animation.

📖 What you need (pages 4, 6, 7 and 8 of this guide)

Law of gravitation: F = G Mm / d², where d is the distance between the centres. Above the surface, d = R + h.
Second law: a = F / m. Near a planet, a = g = GM / R².
Equations of motion: v = u + at, s = ut + ½at², v² = u² + 2as. For a free fall, u = 0 and a = g.
Weight W = mg (unit newton, N). Constants: G = 6.67 × 10⁻¹¹ N m²/kg², M_Earth = 6 × 10²⁴ kg, R_Earth = 6.4 × 10⁶ m, g_Moon = 1.62 m/s².

Problem 1 · A satellite that stops working

N1. 10,000 kg മാസ് ഉള്ള ഒരു കൃത്രിമ ഉപഗ്രഹം പ്രവർത്തനം നിലച്ച് ഭൂമിയിലേക്ക് പതിക്കാൻ തുടങ്ങുന്നു. വസ്തുക്കൾ തമ്മിലുള്ള ആകർഷണം പരസ്പരമാണ് എന്നറിയാമല്ലോ. ഭൂമി ഉപഗ്രഹത്തെ ആകർഷിക്കുന്ന അതേ ബലത്തിൽ ഉപഗ്രഹം ഭൂമിയേയും ആകർഷിക്കുന്നു. ഭൂമിയിൽ നിന്നുള്ള ഉയരം = 5000 m, ഭൂമിയുടെ ആരം R = 6.4 × 10⁶ m An artificial satellite of mass 10,000 kg stops working and starts falling towards the Earth. The attraction between bodies is mutual: the satellite pulls the Earth with the same force with which the Earth pulls the satellite. Height above the Earth = 5000 m, radius of the Earth R = 6.4 × 10⁶ m.

What is the acceleration of the satellite? ഉപഗ്രഹത്തിന്റെ ത്വരണം എത്ര?
What acceleration is produced in the Earth? (mass of the Earth = 6 × 10²⁴ kg) ഭൂമിക്കുണ്ടാകുന്ന ത്വരണം എത്ര?

First find the one force F = GMm/(R + h)² – it is the same on both bodies. Then divide it by each body's own mass: a = F/m for the satellite, a = F/M for the Earth.

🎬 Solution animation – one force, two accelerations

🧪 Try other numbers – raise the satellite, change its mass

height h (km) satellite mass (kg)

Change the mass: the force changes, but the satellite's acceleration does not. Raise the satellite: everything gets smaller, but only slowly at first – 5 km is nothing compared with 6400 km.

Answer: F = GMm/(R + h)² = 6.67 × 10⁻¹¹ × 6 × 10²⁴ × 10⁴ / (6.405 × 10⁶)² ≈ 97550 N (≈ 9.76 × 10⁴ N).
(a) Acceleration of the satellite = F/m = 97550 / 10⁴ ≈ 9.76 m/s² (practically g).
(b) Acceleration of the Earth = F/M = 97550 / (6 × 10²⁴) ≈ 1.6 × 10⁻²⁰ m/s² – so small that the Earth does not move at all for any practical purpose.

⚠️ Common trap

Putting d = 5000 m into the formula. The law uses the distance between the centres, so d = R + h = 6 405 000 m. Using 5000 m would give a force more than a million times too big!

💡 Why this matters

This is the textbook's aeroplane example again: the same force gives a smaller acceleration to a larger mass. That is why we say “the satellite falls to the Earth” and never “the Earth rises to the satellite” – although both pull each other equally hard.

Problem 2 · Dropped from 20 m – Earth and Moon

N2. 10 kg മാസുള്ള ഒരു വസ്തുവിനെ 20 m ഉയരത്തിൽ നിന്ന് ഭൂമിയിലേക്ക് പതിക്കാൻ അനുവദിച്ചാൽ, a) അത് തറയിലെത്താൻ എത്ര സമയമെടുക്കും? b) ചന്ദ്രനിലാണെങ്കിൽ തറയിലെത്താൻ എത്ര സമയം വേണ്ടി വരും എന്ന് കണക്കാക്കുക. (g_ഭൂമി = 10 m/s², g_{ചന്ദ്രൻ} = 1.62 m/s²) An object of mass 10 kg is allowed to fall to the ground from a height of 20 m.

How long does it take to reach the ground?
Calculate how long it would take on the Moon. (g_Earth = 10 m/s², g_Moon = 1.62 m/s²)

“Allowed to fall” means it starts from rest: u = 0. Use s = ut + ½gt², which becomes s = ½gt². Solve for t. Do you need the 10 kg at all?

🎬 Solution animation – two drops

🧪 Try other numbers – a real-time race (Earth vs Moon)

height s (m)

The race runs in real time. Four times the height takes only twice the time – the time grows with the square root of the height.

Answer: u = 0, so s = ½gt² and t = √(2s/g).
(a) Earth: t = √(2 × 20 / 10) = √4 = 2 s.
(b) Moon: t = √(2 × 20 / 1.62) = √24.69 ≈ 4.97 s ≈ 5 s.
The mass, 10 kg, is not needed: the time of fall does not depend on the mass.

⚠️ Common trap

Writing t = 2s/g (forgetting the square root) or t = s/g. Check the unit: s/g has unit m ÷ (m/s²) = s², so you must take a square root to get seconds.

💡 Why this matters

g on the Moon is about 1/6 of the Earth's, but the time is only about √6 ≈ 2.5 times longer, not 6 times. That is why the Apollo astronauts' dropped tools seemed to fall in “slow motion” – but not six times slower.

Problem 3 · A stone thrown up on the Moon

N3. ചന്ദ്രോപരിതലത്തിൽ നിന്ന് ഒരു കല്ല് നേരെ മുകളിലേക്ക് എറിയുന്നു. 6 s കൊണ്ട് കല്ല് തിരിച്ചെത്തുമെങ്കിൽ, a) കല്ലിന്റെ ആദ്യപ്രവേഗം എത്ര? b) കല്ല് സഞ്ചരിച്ചേക്കാവുന്ന ദൂരം എത്ര? c) 4 s കഴിയുമ്പോൾ കല്ലിന്റെ സ്ഥാനം എവിടെയായിരിക്കും? A stone is thrown straight up from the surface of the Moon. If it comes back in 6 s,

what is the initial velocity (ആദ്യപ്രവേഗം) of the stone?
how far can the stone travel (how high does it go)?
where will the stone be after 4 s? (g_Moon = 1.62 m/s²)

Time going up = time coming down, so the stone rises for 3 s. At the very top its velocity is zero. Use v = u − gt for (a), v² = u² − 2gh for (b) and s = ut − ½gt² for (c).

🎬 Solution animation – up for 3 s, down for 3 s

🧪 Try other numbers – follow the stone

return time T (s) time t (s)

The time slider runs from launch (0) to landing (T). Notice that the velocity is + (upwards) for the first half and − (downwards) for the second.

Answer: Time up = 6 ÷ 2 = 3 s.
(a) At the top v = 0: 0 = u − 1.62 × 3, so u = 4.86 m/s upwards.
(b) Maximum height h = u²/2g = 4.86² / (2 × 1.62) = 23.6196 / 3.24 = 7.29 m. (The whole path, up and down, is 2 × 7.29 = 14.58 m.)
(c) s = ut − ½gt² = 4.86 × 4 − ½ × 1.62 × 16 = 19.44 − 12.96 = 6.48 m above the surface, on its way down (it passed the top at 3 s).

⚠️ Common trap

Using t = 6 s for the rise. The stone takes 6 s for the round trip; the rise alone is 3 s. Also keep the sign: going up, gravity slows the stone, so a = −g.

💡 Why this matters

On the Earth the same throw (4.86 m/s) would rise only 4.86²/(2 × 9.8) ≈ 1.2 m and be back in under a second. Weak gravity is why astronauts on the Moon move in long, floating bounds.

Problem 4 · Weight of 10 kg on the Earth and the Moon

N4. 10 കിലോഗ്രാം മാസ് ഉള്ള ഒരു വസ്തുവിന്റെ ഭൂമിയിലെ ഭാരം കണക്കാക്കുക. ഈ വസ്തുവിനെ ചന്ദ്രനിൽ എത്തിച്ചാൽ ഭാരം എത്രയായിരിക്കും? (g_{ചന്ദ്രൻ} = 1.62 m/s²) Calculate the weight on the Earth of an object of mass 10 kg. What will its weight be if it is taken to the Moon? (g_Moon = 1.62 m/s²)

Weight is a force: W = mg. Use g = 9.8 m/s² on the Earth and 1.62 m/s² on the Moon. The mass stays 10 kg in both places.

🎬 Solution animation – the same object on two spring balances

🧪 Try other numbers

mass m (kg)

Answer: Weight on the Earth W = mg = 10 × 9.8 = 98 kg m/s² = 98 N.
Weight on the Moon W = m g_Moon = 10 × 1.62 = 16.2 N.
The mass is 10 kg in both places.

⚠️ Common trap

Writing “weight = 10 kg”. Weight is a force, so its unit is the newton. (The “kg” printed on a spring balance really means kilogram-weight, kgwt – see the next page.)

Check yourself

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