Physics - Mechanics and Properties of Matter - How to Solve: Gravitation (Orbital Velocity, Escape Velocity, Satellite Motion, Intensity, Potential) – NEET UG Physics Guide

How to Solve: Gravitation (Orbital Velocity, Escape Velocity, Satellite Motion, Intensity, Potential) – NEET UG Physics Guide

Introduction

Mastering gravitation unlocks 5-7 marks in NEET UG Physics—enough to push you from a 600 to a 650+ score. These formulas help you predict satellite speeds, calculate escape velocity for rockets, and even understand why planets stay in orbit. If you can solve these problems, you’re not just passing—you’re dominating the exam.

WHAT YOU NEED TO KNOW FIRST

Before diving in, ensure you understand:
1. Newton’s Law of Universal Gravitation – Force between two masses.
2. Circular Motion Basics – Centripetal force, velocity, and acceleration.
3. Potential Energy Concepts – Work done against gravity.

If any of these are unclear, pause and review them first.

KEY TERMS & FORMULAS

1. Orbital Velocity (v₀)

Formula: [ v_0 = \sqrt{\frac{GM}{r}} ] - G = Universal gravitational constant (given on exam sheet) - M = Mass of the central body (e.g., Earth) - r = Radius of the orbit (distance from center of Earth to satellite)

MEMORISE THIS: Orbital velocity is the speed needed for a satellite to stay in a stable circular orbit.

2. Escape Velocity (vₑ)

Formula: [ v_e = \sqrt{\frac{2GM}{r}} ] - Same variables as above.

MEMORISE THIS: Escape velocity is the minimum speed needed to break free from a planet’s gravity without further propulsion.

Key Insight: Escape velocity is √2 times orbital velocity at the same height.

3. Time Period of a Satellite (T)

Formula: [ T = 2\pi \sqrt{\frac{r^3}{GM}} ] - T = Time for one complete orbit (seconds) - r = Radius of orbit (from center of Earth)

MEMORISE THIS: Time period depends only on the radius of the orbit, not the satellite’s mass.

4. Gravitational Potential (V)

Formula: [ V = -\frac{GM}{r} ] - V = Gravitational potential (J/kg) - Negative sign indicates attractive force (potential decreases as you move away).

MEMORISE THIS: Potential is scalar (no direction), unlike force.

5. Gravitational Potential Energy (U)

Formula: [ U = -\frac{GMm}{r} ] - m = Mass of the object (e.g., satellite) - U = Potential energy (Joules)

MEMORISE THIS: Potential energy is always negative because gravity is attractive.

6. Gravitational Intensity (I) / Field Strength (g)

Formula: [ I = \frac{GM}{r^2} ] - I = Gravitational field strength (N/kg or m/s²) - Same as acceleration due to gravity (g) at that point.

MEMORISE THIS: Intensity decreases with square of distance from the center.

STEP-BY-STEP METHOD

Step 1: Identify the Problem Type

• Orbital velocity? → Use ( v_0 = \sqrt{\frac{GM}{r}} )
• Escape velocity? → Use ( v_e = \sqrt{\frac{2GM}{r}} )
• Time period? → Use ( T = 2\pi \sqrt{\frac{r^3}{GM}} )
• Potential or energy? → Use ( V = -\frac{GM}{r} ) or ( U = -\frac{GMm}{r} )
• Intensity/field strength? → Use ( I = \frac{GM}{r^2} )

Step 2: List Given Values

• Mass of Earth (M) = ( 6 \times 10^{24} ) kg (given on exam sheet)
• Radius of Earth (R) = ( 6.4 \times 10^6 ) m (given on exam sheet)
• G = ( 6.67 \times 10^{-11} ) Nm²/kg² (given on exam sheet)
• Height (h) = Given in problem (if satellite is at height h, then r = R + h)

Step 3: Plug into the Correct Formula

• Substitute values carefully.
• Check units: Ensure all values are in SI units (kg, m, s).
• Simplify step-by-step to avoid calculation errors.

Step 4: Solve & Verify

• Calculate the final value.
• Cross-check: Does the answer make sense?
• Orbital velocity should be ~7.9 km/s near Earth’s surface.
• Escape velocity should be ~11.2 km/s near Earth’s surface.

WORKED EXAMPLES

Example 1 – Basic (Orbital Velocity)

Problem: A satellite orbits Earth at a height of 300 km. Find its orbital velocity. Given: - ( M = 6 \times 10^{24} ) kg - ( R = 6.4 \times 10^6 ) m - ( h = 300 \times 10^3 ) m - ( G = 6.67 \times 10^{-11} ) Nm²/kg²

Solution:
1. Find r (radius of orbit): [ r = R + h = 6.4 \times 10^6 + 300 \times 10^3 = 6.7 \times 10^6 \, \text{m} ]

1. Use orbital velocity formula: [ v_0 = \sqrt{\frac{GM}{r}} ] [ v_0 = \sqrt{\frac{(6.67 \times 10^{-11})(6 \times 10^{24})}{6.7 \times 10^6}} ]

2. Calculate inside the square root: [ \frac{(6.67 \times 6) \times 10^{13}}{6.7 \times 10^6} = \frac{40.02 \times 10^{13}}{6.7 \times 10^6} \approx 5.97 \times 10^7 ]

3. Take square root: [ v_0 = \sqrt{5.97 \times 10^7} \approx 7.73 \times 10^3 \, \text{m/s} ] [ v_0 \approx 7.73 \, \text{km/s} ]

What we did and why: - We found the total radius (r) by adding Earth’s radius and the satellite’s height. - Used the orbital velocity formula because the problem asked for speed in orbit. - Verified that 7.73 km/s is reasonable (close to Earth’s surface orbital speed of 7.9 km/s).

Example 2 – Medium (Escape Velocity)

Problem: What is the escape velocity from the surface of the Moon? Given: - Mass of Moon (M) = ( 7.35 \times 10^{22} ) kg - Radius of Moon (R) = ( 1.74 \times 10^6 ) m - ( G = 6.67 \times 10^{-11} ) Nm²/kg²

Solution:
1. Escape velocity formula: [ v_e = \sqrt{\frac{2GM}{R}} ]

1. Substitute values: [ v_e = \sqrt{\frac{2 \times 6.67 \times 10^{-11} \times 7.35 \times 10^{22}}{1.74 \times 10^6}} ]

2. Calculate numerator: [ 2 \times 6.67 \times 7.35 \times 10^{11} \approx 98.1 \times 10^{11} ]

3. Divide by denominator: [ \frac{98.1 \times 10^{11}}{1.74 \times 10^6} \approx 5.64 \times 10^6 ]

4. Take square root: [ v_e = \sqrt{5.64 \times 10^6} \approx 2.37 \times 10^3 \, \text{m/s} ] [ v_e \approx 2.37 \, \text{km/s} ]

What we did and why: - Used the escape velocity formula because the problem asked for the minimum speed to escape the Moon’s gravity. - Noted that 2.37 km/s is much lower than Earth’s escape velocity (11.2 km/s), which makes sense because the Moon is less massive.

Example 3 – Exam-Style (Time Period & Energy)

Problem: A satellite of mass 500 kg orbits Earth at a height of 400 km. (a) Find its orbital time period. (b) Calculate its gravitational potential energy.

Given: - ( M = 6 \times 10^{24} ) kg - ( R = 6.4 \times 10^6 ) m - ( h = 400 \times 10^3 ) m - ( m = 500 ) kg - ( G = 6.67 \times 10^{-11} ) Nm²/kg²

Solution (a): Time Period
1. Find r: [ r = R + h = 6.4 \times 10^6 + 400 \times 10^3 = 6.8 \times 10^6 \, \text{m} ]

1. Use time period formula: [ T = 2\pi \sqrt{\frac{r^3}{GM}} ]

2. Calculate ( r^3 ): [ (6.8 \times 10^6)^3 = 314.4 \times 10^{18} ]

3. Divide by GM: [ \frac{314.4 \times 10^{18}}{6.67 \times 6 \times 10^{13}} = \frac{314.4}{40.02} \times 10^5 \approx 7.86 \times 10^5 ]

4. Take square root & multiply by ( 2\pi ): [ \sqrt{7.86 \times 10^5} \approx 886.5 ] [ T = 2\pi \times 886.5 \approx 5570 \, \text{s} ] [ T \approx 92.8 \, \text{minutes} ]

Solution (b): Potential Energy
1. Use potential energy formula: [ U = -\frac{GMm}{r} ]

1. Substitute values: [ U = -\frac{(6.67 \times 10^{-11})(6 \times 10^{24})(500)}{6.8 \times 10^6} ]

2. Calculate numerator: [ 6.67 \times 6 \times 500 \times 10^{13} = 20 \times 10^{16} ]

3. Divide by r: [ \frac{20 \times 10^{16}}{6.8 \times 10^6} \approx 2.94 \times 10^{10} ]

4. Final answer: [ U = -2.94 \times 10^{10} \, \text{J} ]

What we did and why: - For (a), we used the time period formula because the question asked for how long one orbit takes. - For (b), we used potential energy formula because the question asked for energy, not just potential. - Verified that 92.8 minutes is reasonable (close to the 90-minute low Earth orbit). - Noted that potential energy is negative, confirming gravity is attractive.

COMMON MISTAKES

EXAM TRAPS

1-MINUTE RECAP (Night Before Exam)

"Listen up—this is your 60-second gravitation cheat sheet for NEET!

1. Orbital velocity = ( \sqrt{\frac{GM}{r}} ). Escape velocity = ( \sqrt{\frac{2GM}{r}} ). Remember: Escape is √2 times orbital speed.
2. Time period = ( 2\pi \sqrt{\frac{r^3}{GM}} ). Only depends on radius, not mass!
3. Potential (V) = ( -\frac{GM}{r} ). Potential energy (U) = ( -\frac{GMm}{r} ). Negative sign is crucial!
4. Intensity (g) = ( \frac{GM}{r^2} ). Same as acceleration due to gravity at that point.
5. Always add height to Earth’s radius (r = R + h). Never use height alone!
6. Units matter! Convert km to meters, grams to kg.
7. If stuck, compare ratios. For example, if radius doubles, orbital velocity becomes ( \frac{1}{\sqrt{2}} ) times.

You’ve got this. Now go crush those gravitation questions!