Problem 1
EASY
Two masses of 1000 kg and 2000 kg are placed 10 m apart. Calculate the gravitational force between them.
Given:
\(m_1 = 1000\) kg
\(m_2 = 2000\) kg
\(r = 10\) m
\(G = 6.67 \times 10^{-11}\) N⋅m²/kg²
\(m_1 = 1000\) kg
\(m_2 = 2000\) kg
\(r = 10\) m
\(G = 6.67 \times 10^{-11}\) N⋅m²/kg²
Step-by-Step Solution:
Step 1: Write the formula
\[F = G\frac{m_1 m_2}{r^2}\]
Step 2: Substitute the values
\[F = 6.67 \times 10^{-11} \times \frac{1000 \times 2000}{10^2}\]
Step 3: Calculate
\[F = 6.67 \times 10^{-11} \times \frac{2 \times 10^6}{100}\]
\[F = 6.67 \times 10^{-11} \times 2 \times 10^4\]
\[F = 1.334 \times 10^{-6} \text{ N}\]
\[\boxed{F = 1.334 \times 10^{-6} \text{ N}}\]
Problem 2
EASY
Calculate the gravitational force between Earth and a 70 kg person standing on its surface.
Given:
\(m_1 = 70\) kg (person)
\(M_E = 5.97 \times 10^{24}\) kg (Earth)
\(R_E = 6.37 \times 10^6\) m
\(G = 6.67 \times 10^{-11}\) N⋅m²/kg²
\(m_1 = 70\) kg (person)
\(M_E = 5.97 \times 10^{24}\) kg (Earth)
\(R_E = 6.37 \times 10^6\) m
\(G = 6.67 \times 10^{-11}\) N⋅m²/kg²
Step 1: Apply Newton’s law
\[F = G\frac{m \cdot M_E}{R_E^2}\]
Step 2: Substitute values
\[F = 6.67 \times 10^{-11} \times \frac{70 \times 5.97 \times 10^{24}}{(6.37 \times 10^6)^2}\]
Step 3: Calculate
\[F = 6.67 \times 10^{-11} \times \frac{4.179 \times 10^{26}}{4.06 \times 10^{13}}\]
\[F = 6.67 \times 10^{-11} \times 1.029 \times 10^{13}\]
\[F = 686 \text{ N}\]
\[\boxed{F = 686 \text{ N (Weight of the person)}}\]
Problem 3
MEDIUM
Two spheres of masses 500 kg and 800 kg experience a gravitational force of \(2.67 \times 10^{-6}\) N. Find the distance between their centers.
Given:
\(m_1 = 500\) kg
\(m_2 = 800\) kg
\(F = 2.67 \times 10^{-6}\) N
\(G = 6.67 \times 10^{-11}\) N⋅m²/kg²
Find: \(r = ?\)
\(m_1 = 500\) kg
\(m_2 = 800\) kg
\(F = 2.67 \times 10^{-6}\) N
\(G = 6.67 \times 10^{-11}\) N⋅m²/kg²
Find: \(r = ?\)
Step 1: Rearrange the formula for r
\[F = G\frac{m_1 m_2}{r^2} \Rightarrow r^2 = G\frac{m_1 m_2}{F}\]
\[r = \sqrt{G\frac{m_1 m_2}{F}}\]
Step 2: Substitute values
\[r = \sqrt{6.67 \times 10^{-11} \times \frac{500 \times 800}{2.67 \times 10^{-6}}}\]
Step 3: Calculate
\[r = \sqrt{6.67 \times 10^{-11} \times \frac{4 \times 10^5}{2.67 \times 10^{-6}}}\]
\[r = \sqrt{6.67 \times 10^{-11} \times 1.498 \times 10^{11}}\]
\[r = \sqrt{100} = 10 \text{ m}\]
\[\boxed{r = 10 \text{ m}}\]
Problem 4
EASY
Calculate the gravitational force between two 1500 kg cars parked 5 m apart.
Given:
\(m_1 = m_2 = 1500\) kg
\(r = 5\) m
\(G = 6.67 \times 10^{-11}\) N⋅m²/kg²
\(m_1 = m_2 = 1500\) kg
\(r = 5\) m
\(G = 6.67 \times 10^{-11}\) N⋅m²/kg²
Step 1: Apply the formula
\[F = G\frac{m_1 m_2}{r^2} = G\frac{m^2}{r^2}\]
Step 2: Substitute values
\[F = 6.67 \times 10^{-11} \times \frac{(1500)^2}{5^2}\]
\[F = 6.67 \times 10^{-11} \times \frac{2.25 \times 10^6}{25}\]
Step 3: Calculate
\[F = 6.67 \times 10^{-11} \times 9 \times 10^4\]
\[F = 6.003 \times 10^{-6} \text{ N}\]
\[\boxed{F = 6.003 \times 10^{-6} \text{ N}}\]
Problem 5
MEDIUM
A satellite of mass 1000 kg orbits Earth at an altitude of 400 km above the surface. Calculate the gravitational force acting on it.
Given:
\(m = 1000\) kg
\(h = 400\) km = \(4 \times 10^5\) m
\(M_E = 5.97 \times 10^{24}\) kg
\(R_E = 6.37 \times 10^6\) m
Distance from center: \(r = R_E + h\)
\(m = 1000\) kg
\(h = 400\) km = \(4 \times 10^5\) m
\(M_E = 5.97 \times 10^{24}\) kg
\(R_E = 6.37 \times 10^6\) m
Distance from center: \(r = R_E + h\)
Step 1: Calculate total distance from Earth’s center
\[r = R_E + h = 6.37 \times 10^6 + 4 \times 10^5 = 6.77 \times 10^6 \text{ m}\]
Step 2: Apply gravitational law
\[F = G\frac{m \cdot M_E}{r^2}\]
\[F = 6.67 \times 10^{-11} \times \frac{1000 \times 5.97 \times 10^{24}}{(6.77 \times 10^6)^2}\]
Step 3: Calculate
\[F = 6.67 \times 10^{-11} \times \frac{5.97 \times 10^{27}}{4.58 \times 10^{13}}\]
\[F = 6.67 \times 10^{-11} \times 1.30 \times 10^{14}\]
\[F = 8671 \text{ N}\]
\[\boxed{F = 8671 \text{ N}}\]
Problem 6
HARD
Two identical spheres each of mass M are placed such that the gravitational force between them is F. If the distance between them is doubled, by what factor does the force change?
Given:
Initial masses: \(m_1 = m_2 = M\)
Initial distance: \(r_1\)
Initial force: \(F_1 = F\)
Final distance: \(r_2 = 2r_1\)
Find: \(\frac{F_2}{F_1}\)
Initial masses: \(m_1 = m_2 = M\)
Initial distance: \(r_1\)
Initial force: \(F_1 = F\)
Final distance: \(r_2 = 2r_1\)
Find: \(\frac{F_2}{F_1}\)
Step 1: Write initial force
\[F_1 = G\frac{M^2}{r_1^2}\]
Step 2: Write final force when distance is doubled
\[F_2 = G\frac{M^2}{(2r_1)^2} = G\frac{M^2}{4r_1^2}\]
Step 3: Find the ratio
\[\frac{F_2}{F_1} = \frac{G\frac{M^2}{4r_1^2}}{G\frac{M^2}{r_1^2}} = \frac{1}{4}\]
\[\boxed{\text{Force becomes } \frac{1}{4} \text{ of the original value}}\]
General Rule: Force is inversely proportional to the square of distance
Problem 7
EASY
Find the gravitational force between a 50 kg student and a 60 kg teacher standing 2 m apart.
Given:
\(m_1 = 50\) kg
\(m_2 = 60\) kg
\(r = 2\) m
\(m_1 = 50\) kg
\(m_2 = 60\) kg
\(r = 2\) m
Solution:
\[F = 6.67 \times 10^{-11} \times \frac{50 \times 60}{2^2}\]
\[F = 6.67 \times 10^{-11} \times \frac{3000}{4}\]
\[F = 6.67 \times 10^{-11} \times 750\]
\[F = 5.0025 \times 10^{-8} \text{ N}\]
\[\boxed{F = 5.00 \times 10^{-8} \text{ N}}\]
Problem 8
MEDIUM
Calculate the gravitational acceleration on the surface of Mars. (Mass of Mars = \(6.39 \times 10^{23}\) kg, Radius = \(3.39 \times 10^6\) m)
Given:
\(M_{Mars} = 6.39 \times 10^{23}\) kg
\(R_{Mars} = 3.39 \times 10^6\) m
Find: \(g_{Mars}\)
\(M_{Mars} = 6.39 \times 10^{23}\) kg
\(R_{Mars} = 3.39 \times 10^6\) m
Find: \(g_{Mars}\)
Formula for surface gravity:
\[g = \frac{GM}{R^2}\]
Substitute values:
\[g_{Mars} = \frac{6.67 \times 10^{-11} \times 6.39 \times 10^{23}}{(3.39 \times 10^6)^2}\]
\[g_{Mars} = \frac{4.26 \times 10^{13}}{1.15 \times 10^{13}}\]
\[g_{Mars} = 3.71 \text{ m/s}^2\]
\[\boxed{g_{Mars} = 3.71 \text{ m/s}^2}\]
(About 38% of Earth’s gravity)
Problem 9
HARD
At what height above Earth’s surface will the gravitational acceleration be 25% of its surface value?
Given:
\(g_h = 0.25g\) (where \(g = 9.8\) m/s²)
\(R_E = 6.37 \times 10^6\) m
Find: Height \(h\)
\(g_h = 0.25g\) (where \(g = 9.8\) m/s²)
\(R_E = 6.37 \times 10^6\) m
Find: Height \(h\)
Step 1: Write gravity at height h
\[g_h = \frac{GM}{(R_E + h)^2}\]
\[g = \frac{GM}{R_E^2}\] (at surface)
Step 2: Set up the ratio
\[\frac{g_h}{g} = \frac{R_E^2}{(R_E + h)^2} = 0.25 = \frac{1}{4}\]
Step 3: Solve for h
\[\frac{R_E^2}{(R_E + h)^2} = \frac{1}{4}\]
\[(R_E + h)^2 = 4R_E^2\]
\[R_E + h = 2R_E\]
\[h = R_E = 6.37 \times 10^6 \text{ m}\]
\[\boxed{h = 6370 \text{ km (One Earth radius)}}\]
Problem 10
MEDIUM
Two masses \(m_1 = 100\) kg and \(m_2 = 200\) kg are separated by 4 m. Find the point where the gravitational field due to both masses is zero.
Given:
\(m_1 = 100\) kg, \(m_2 = 200\) kg
Separation = 4 m
Find: Neutral point location
\(m_1 = 100\) kg, \(m_2 = 200\) kg
Separation = 4 m
Find: Neutral point location
Step 1: Let neutral point be at distance x from \(m_1\)
Distance from \(m_2\) = \((4-x)\) m
Step 2: At neutral point, gravitational fields cancel
\[\frac{Gm_1}{x^2} = \frac{Gm_2}{(4-x)^2}\]
\[\frac{100}{x^2} = \frac{200}{(4-x)^2}\]
Step 3: Solve the equation
\[\frac{(4-x)^2}{x^2} = \frac{200}{100} = 2\]
\[\frac{4-x}{x} = \sqrt{2} = 1.414\]
\[4-x = 1.414x\]
\[4 = 2.414x\]
\[x = 1.66 \text{ m}\]
\[\boxed{\text{Neutral point is 1.66 m from the 100 kg mass}}\]
Problem 11
EASY
Calculate the weight of a 80 kg astronaut on the Moon’s surface. (Moon’s mass = \(7.35 \times 10^{22}\) kg, radius = \(1.74 \times 10^6\) m)
Given:
\(m = 80\) kg
\(M_{Moon} = 7.35 \times 10^{22}\) kg
\(R_{Moon} = 1.74 \times 10^6\) m
\(m = 80\) kg
\(M_{Moon} = 7.35 \times 10^{22}\) kg
\(R_{Moon} = 1.74 \times 10^6\) m
Step 1: Find Moon’s surface gravity
\[g_{Moon} = \frac{GM_{Moon}}{R_{Moon}^2}\]
\[g_{Moon} = \frac{6.67 \times 10^{-11} \times 7.35 \times 10^{22}}{(1.74 \times 10^6)^2}\]
\[g_{Moon} = 1.62 \text{ m/s}^2\]
Step 2: Calculate weight
\[W = mg_{Moon} = 80 \times 1.62 = 129.6 \text{ N}\]
\[\boxed{W = 130 \text{ N (approximately 1/6 of Earth weight)}}\]
Problem 12
MEDIUM
If the mass of one object is tripled and the distance between two objects is halved, by what factor does the gravitational force change?
Initial force:
\[F_1 = G\frac{m_1m_2}{r^2}\]
Final force:
\[F_2 = G\frac{3m_1 \cdot m_2}{(r/2)^2} = G\frac{3m_1m_2}{r^2/4} = 12G\frac{m_1m_2}{r^2}\]
Factor of change:
\[\frac{F_2}{F_1} = \frac{12G\frac{m_1m_2}{r^2}}{G\frac{m_1m_2}{r^2}} = 12\]
\[\boxed{\text{Force increases by a factor of 12}}\]
Problem 13
EASY
Two point masses of 300 kg and 400 kg are 6 m apart. Calculate the gravitational force between them.
\[F = 6.67 \times 10^{-11} \times \frac{300 \times 400}{6^2}\]
\[F = 6.67 \times 10^{-11} \times \frac{120000}{36}\]
\[F = 6.67 \times 10^{-11} \times 3333.33\]
\[F = 2.22 \times 10^{-7} \text{ N}\]
\[\boxed{F = 2.22 \times 10^{-7} \text{ N}}\]
Problem 14
HARD
Calculate the escape velocity from Earth’s surface using gravitational principles.
Energy conservation principle:
\[\frac{1}{2}mv_e^2 – \frac{GMm}{R} = 0\]
(Kinetic energy = Gravitational potential energy)
Solve for escape velocity:
\[v_e = \sqrt{\frac{2GM}{R}} = \sqrt{2gR}\]
\[v_e = \sqrt{2 \times 9.8 \times 6.37 \times 10^6}\]
\[v_e = \sqrt{1.25 \times 10^8} = 11180 \text{ m/s}\]
\[\boxed{v_e = 11.18 \text{ km/s}}\]
Problem 15
MEDIUM
A 500 kg satellite orbits Earth at a height where gravitational acceleration is 8 m/s². Find the orbital height.
\[g_h = \frac{GM}{(R+h)^2} = 8 \text{ m/s}^2\]
\[g = \frac{GM}{R^2} = 9.8 \text{ m/s}^2\]
\[\frac{g_h}{g} = \frac{R^2}{(R+h)^2} = \frac{8}{9.8} = 0.816\]
\[(R+h)^2 = \frac{R^2}{0.816}\]
\[R+h = \frac{R}{\sqrt{0.816}} = \frac{R}{0.904}\]
\[h = R(1.106 – 1) = 0.106R\]
\[h = 0.106 \times 6.37 \times 10^6 = 6.75 \times 10^5 \text{ m}\]
\[\boxed{h = 675 \text{ km}}\]
Problem 16
EASY
Find gravitational force between 250 kg and 750 kg masses 8 m apart.
\[F = 6.67 \times 10^{-11} \times \frac{250 \times 750}{64} = 1.95 \times 10^{-7} \text{ N}\]
Problem 17
MEDIUM
Calculate gravity on Jupiter’s surface. (Mass = \(1.9 \times 10^{27}\) kg, Radius = \(7.15 \times 10^7\) m)
\[g_J = \frac{6.67 \times 10^{-11} \times 1.9 \times 10^{27}}{(7.15 \times 10^7)^2} = 24.8 \text{ m/s}^2\]
Problem 18
EASY
Two 900 kg masses are 12 m apart. Find the gravitational attraction.
\[F = 6.67 \times 10^{-11} \times \frac{900^2}{144} = 3.75 \times 10^{-6} \text{ N}\]
Problem 19
HARD
Find the ratio of gravitational forces on a mass at Earth’s surface vs. at Moon’s surface.
\[\frac{F_E}{F_M} = \frac{g_E}{g_M} = \frac{9.8}{1.62} = 6.05\]
Earth’s gravity is about 6 times stronger than Moon’s.
Problem 20
MEDIUM
A 200 kg object experiences 1960 N gravitational force at Earth’s surface. Verify the calculation using Newton’s law.
\[F = mg = 200 \times 9.8 = 1960 \text{ N} \checkmark\]
\[F = \frac{GMm}{R^2} = \frac{6.67 \times 10^{-11} \times 5.97 \times 10^{24} \times 200}{(6.37 \times 10^6)^2} = 1960 \text{ N} \checkmark\]
🎯 Exam Tips & Key Points
🔢 Formula Memory:
\(F = G\frac{m_1m_2}{r^2}\)
\(G = 6.67 \times 10^{-11}\) N⋅m²/kg²
\(F = G\frac{m_1m_2}{r^2}\)
\(G = 6.67 \times 10^{-11}\) N⋅m²/kg²
📐 Distance Rule:
Force ∝ \(\frac{1}{r^2}\)
Double distance → Force becomes 1/4
Force ∝ \(\frac{1}{r^2}\)
Double distance → Force becomes 1/4
⚖️ Mass Rule:
Force ∝ \(m_1 \times m_2\)
Double mass → Double force
Force ∝ \(m_1 \times m_2\)
Double mass → Double force
🌍 Surface Gravity:
\(g = \frac{GM}{R^2}\)
Weight = mg
\(g = \frac{GM}{R^2}\)
Weight = mg
🛰️ At Height h:
\(g_h = g\left(\frac{R}{R+h}\right)^2\)
Gravity decreases with altitude
\(g_h = g\left(\frac{R}{R+h}\right)^2\)
Gravity decreases with altitude
⚡ Quick Calculations:
Use scientific notation
Round G to 6.7×10⁻¹¹ for speed
Use scientific notation
Round G to 6.7×10⁻¹¹ for speed