SaitechAI Worksheet — Gravitation (Example + Practice)

Strategy to solve numericals: 1) Problem2) Given3) To find4) Formula5) Substitute6) Final answer with SI units & significant figures

Example Universal Gravitation (Finding distance)

1. Problem
A sphere of mass 40 kg is attracted by a second sphere of mass 60 kg with a force equal to 4 mgf. If G = 6×10⁻¹¹ N·m²·kg⁻² and g = 10 m·s⁻², find the distance between them.
2. Given parameters (symbols & SI units)
M = 40 kg, m = 60 kg, G = 6×10⁻¹¹ N·m²·kg⁻², g = 10 m·s⁻²
Force given: F = 4 mgf (here mgf = (10⁻⁶ kg)·g)
F = 4×(10⁻⁶ kg)×(10 m·s⁻²) = 4×10⁻⁵ N
3. To find
Distance r between the spheres (in metres).
4. Formula applicable
Newton’s law of gravitation: F = (G M m) / r²
5. Apply the parameters
Rearranging: r = √(G M m / F)
r = √[(6×10⁻¹¹ × 40 × 60) / (4×10⁻⁵)]
6. Solution (SI units & sig. figs.)
r = √(3.6×10⁻³) = 0.060 m
Answer: r = 0.060 m = 6.0 cm

Tip: Always keep r in metres in the final SI answer. Convert to cm only as an additional step.

Student Details

Use G = 6.67×10⁻¹¹ N·m²·kg⁻² unless a different value is given in the question.

Worksheet 6 Similar Questions (Attempt first, then Submit)

Q1
2 marks

Two spheres of masses 20 kg and 50 kg attract each other with force F = 1.0×10⁻⁶ N. Take G = 6.67×10⁻¹¹ N·m²·kg⁻². Find the distance r between them.

Q2
2 marks

Two spheres each of mass 10 kg are separated by 0.50 m. Using G = 6.67×10⁻¹¹ (SI), find the gravitational force between them.

Q3
2 marks

A mass m = 5.0 kg is at a distance r = 0.20 m from another mass M = 30 kg. Take G = 6.67×10⁻¹¹ (SI). Find the force of attraction.

Q4
2 marks

Two masses M = 80 kg and m = 20 kg attract with force F = 5.0×10⁻⁷ N. Take G = 6.67×10⁻¹¹ (SI). Find their separation r.

Q5
2 marks

At what distance from a 60 kg mass will the gravitational field (acceleration due to that mass) be g = 1.0×10⁻⁸ m·s⁻²? Take G = 6.67×10⁻¹¹ (SI). (Use g = GM/r².)

Q6
2 marks

Two masses 40 kg and 60 kg are placed 0.30 m apart. Find the gravitational force between them using G = 6.67×10⁻¹¹ (SI).

Answer Key (shown after Submit)

Use SI units. Answers shown with reasonable significant figures.

Q1

r = √(G M m / F)
r = √[(6.67×10⁻¹¹ × 20 × 50) / (1.0×10⁻⁶)]
r ≈ 0.258 m

Q2

F = G M m / r²
F = 6.67×10⁻¹¹ × 10 × 10 / (0.50)²
F ≈ 2.67×10⁻⁸ N

Q3

F = G M m / r²
F = 6.67×10⁻¹¹ × 30 × 5.0 / (0.20)²
F ≈ 2.50×10⁻⁷ N

Q4

r = √(G M m / F)
r = √[(6.67×10⁻¹¹ × 80 × 20) / (5.0×10⁻⁷)]
r ≈ 0.462 m

Q5

g = GM / r² ⇒ r = √(GM / g)
r = √[(6.67×10⁻¹¹ × 60) / (1.0×10⁻⁸)]
r ≈ 0.632 m

Q6

F = G M m / r²
F = 6.67×10⁻¹¹ × 40 × 60 / (0.30)²
F ≈ 1.78×10⁻⁶ N

After Submit, compare your final SI answer and your substitution steps. If your magnitude differs, recheck powers of 10 and unit conversions.