Gravitation | ISC Class 11 Physics Notes

This note covers Kepler’s laws, universal gravitation, gravitational field and potential, acceleration due to gravity and its variation, gravitational potential energy, escape speed, satellite motion, weightlessness, geostationary orbits and polar satellites.

What does Newton’s universal law of gravitation describe?

Gravitation is the mutual attraction between masses. Newton’s universal law relates the strength of this attraction to the masses and their separation. It connects the fall of objects towards Earth with the motion of the Moon and planets.

Definition: Every body attracts every other body with a force proportional to the product of their masses and inversely proportional to the square of the distance between them.

For two point masses, whose sizes can be neglected for the calculation, let m₁ and m₂ denote their masses and r their separation. Let F denote the magnitude of the gravitational force and G the universal gravitational constant.

F = Gm₁m₂/r²

The force acts along the line joining the masses and is attractive. The two bodies exert equal and opposite forces on each other. Equal forces need not produce equal accelerations because acceleration is force divided by the mass being accelerated.

How are units and dimensions assigned?

The International System of Units, abbreviated SI, uses kilogram (kg) for mass, metre (m) for length and second (s) for time. The SI unit of force is the newton (N), with 1 N = 1 kg m s⁻².

The SI unit of G is N m² kg⁻². Use G = 6.67 × 10⁻¹¹ N m² kg⁻² for calculations here. Rearranging the force equation gives G = Fr²/(m₁m₂).

In dimensional notation, M, L and T represent mass, length and time respectively; square brackets mean “dimensions of”. Thus [G] = [M⁻¹L³T⁻²]. These dimensional letters differ from symbols for particular physical quantities used later.

When can an extended body be treated as a point mass?

Outside a sphere with a spherically symmetric mass distribution, its gravitational attraction is the same as if its mass were concentrated at its centre. Spherical symmetry means that the mass distribution is unchanged by rotation about that centre.

Inside a uniform hollow spherical shell, the force due to that shell is zero. For several attracting masses, use superposition: calculate their forces independently and add them as vectors, quantities with magnitude and direction. Adding force magnitudes alone generally gives an incorrect resultant.

How do Kepler’s laws explain planetary motion?

An orbit is the path followed by a body revolving around another body. Kepler’s three laws describe planetary orbits, the changing rate of motion along them, and the relationship between orbital size and the time taken for a revolution.

What do the three laws state?

1. Law of orbits: Planets move in elliptical orbits with the Sun at one focus. An ellipse is a closed curve for which the sum of distances from two fixed points, called foci, remains constant.
2. Law of areas: The line joining the Sun and a planet sweeps equal areas in equal intervals of time. The planet moves faster nearer the Sun and slower farther away.
3. Law of periods: The square of the time period of revolution is proportional to the cube of the semi-major axis of the orbit.

Let T now denote the time period, the time for one complete revolution, and a the semi-major axis, half the longest diameter of the ellipse. Then T² ∝ a³, or T²/a³ is constant for planets orbiting the same central body.

For a circular orbit, the semi-major axis equals the radius. The constant in the period law depends on the central mass, so a value for planets orbiting the Sun cannot simply be used for satellites orbiting Earth.

The ellipse contains two foci, S and S′, with the Sun at S. P is the closest point, or perihelion, and A the farthest point, or aphelion. The full major axis is labelled 2a; the full minor axis is labelled 2b, where b is the semi-minor axis.

See Fig. 7.1(a) in your NCERT textbook

Why is the swept area constant per unit time?

A central force acts along the line joining the moving body to a fixed centre. Gravity is central in the planetary model. Its torque, or turning effect about the Sun, is zero because the force acts along this line.

Consequently angular momentum, the moment about the Sun of linear momentum, which is mass multiplied by velocity, is conserved. If L denotes its magnitude and m the planet’s mass, the area swept per unit time is L/(2m). A constant angular momentum therefore gives Kepler’s area law.

At perihelion and aphelion, velocity is perpendicular to the Sun-planet line. Writing the distances as rₚ and rₐ and speeds as vₚ and vₐ gives rₚvₚ = rₐvₐ. A smaller perihelion distance therefore corresponds to a larger speed.

How are gravitational field, acceleration and weight related?

A gravitational field is the region in which a mass experiences gravitational force due to another mass. The field intensity at a point is the gravitational force per unit test mass placed there. A test mass is small enough not to disturb the source appreciably.

Let I denote field intensity and m the test mass. In vector form, I = F/m, where F is the force vector. The SI unit of gravitational field intensity is N kg⁻¹, equivalent to m s⁻².

Acceleration due to gravity is the acceleration produced by gravitational attraction. Newton’s second law gives F = mg, so gravitational field intensity and gravitational acceleration have the same magnitude and direction. Here g denotes the acceleration at the position being considered.

What is the surface value for a spherical Earth?

Let M be Earth’s mass and R its radius. In the spherical, non-rotating model, the surface force is GMm/R². Dividing by the falling mass gives g = GM/R². The SI unit of acceleration due to gravity is m s⁻².

The weight W of a body is the gravitational force on it: W = mg. The SI unit of weight is N. Mass is a property of the body; weight changes when the local gravitational acceleration changes.

The cancellation of m explains why the gravitational acceleration does not depend on the falling body’s mass. This statement concerns motion under gravity; air resistance can affect the observed motion of bodies differently.

Worked example 1. Estimate Earth’s mass using surface acceleration g = 9.81 m s⁻², radius R = 6.37 × 10⁶ m and G = 6.67 × 10⁻¹¹ N m² kg⁻².

Formula: g = GM/R²; M = gR²/G.

Substitute: M = 9.81 × (6.37 × 10⁶)²/(6.67 × 10⁻¹¹).

Answer: M ≈ 5.97 × 10²⁴ kg. Measuring surface gravity and radius therefore allows the mass to be inferred once G is known.

Keep G and g distinct. G is the universal constant in the gravitational force law, whereas g is a field acceleration that depends on the source mass and position. They have different units and are not interchangeable.

How does gravity change with altitude?

Altitude h is height above Earth’s surface. For an object at altitude h, the distance from Earth’s centre is R + h, not h. This distinction is essential because the inverse-square force law uses separation from the centre of the spherical source.

Derivation: Gravity above the surface

1. Take Earth to be spherical, with mass M and radius R, and neglect its rotation. A body of mass m at altitude h experiences force F = GMm/(R + h)².
2. Divide by m to obtain the acceleration at altitude h, denoted by gₕ: gₕ = GM/(R + h)².
3. Use the surface relation g = GM/R² and divide the two expressions to eliminate GM.

gₕ = g[R/(R + h)]²

This result is valid at any external altitude within the spherical model. Gravity decreases as the distance from the centre increases. It does not suddenly become zero when an object reaches space.

When is the small-height approximation valid?

If h ≪ R, meaning h is much smaller than Earth’s radius, a binomial expansion gives gₕ ≈ g(1 − 2h/R). The symbol ≈ means approximately equal. Terms containing higher powers of h/R are neglected.

The approximation must not be used when the height is comparable with R. The exact ratio remains straightforward to use in such cases and avoids obtaining an unphysical result from an approximation applied outside its range.

A shaded circle represents Earth. Its surface is labelled, a radius inside the circle is marked Rₑ, meaning Earth’s radius, and a vertical segment above the surface is labelled h. The centre-to-object distance is therefore Rₑ + h.

See Fig. 7.8(a) in your NCERT textbook

Worked example 2. A body weighs 63 N on Earth’s surface. Find its gravitational force at a height equal to half Earth’s radius, treating Earth as spherical.

Formula: Wₕ/W = [R/(R + h)]², where Wₕ is weight at altitude and W is surface weight.

Substitute: h = R/2, so Wₕ = 63 × [R/(3R/2)]² = 63 × 4/9.

Answer: Wₕ = 28 N. Its mass has not changed; the gravitational field is weaker.

How do depth, latitude and rotation affect gravity?

Depth d is distance below Earth’s surface. In the ideal model of a sphere of uniform density, the mass within a smaller concentric sphere is proportional to the cube of its radius. Uniform density means equal mass per unit volume throughout the sphere.

Derivation: Gravity below the surface

1. A point at depth d has radial distance r = R − d from Earth’s centre. Let Mᵣ denote the mass enclosed within radius r.
2. Uniform density gives Mᵣ/M = r³/R³. The outer spherical shells produce zero net gravitational force at the point.
3. Thus the local acceleration, denoted by gᵈ, is GMᵣ/r² = GMr/R³. Substitute g = GM/R² and r = R − d.

gᵈ = g(1 − d/R)

Under this assumption gravity decreases linearly with depth, reaching zero at the centre. Do not describe this straight-line result as a calculation using Earth’s actual internal density distribution.

An outer circle marks Earth’s surface and an inner dashed circle encloses a smaller sphere. The radial separation of their surfaces is d. The labels Mₑ and Mₛ denote Earth’s mass and the enclosed smaller mass respectively.

See Fig. 7.8(b) in your NCERT textbook

For a uniform, non-rotating sphere, put radial distance on the horizontal axis and gravitational acceleration on the vertical axis. Draw a straight rise from zero at the centre to g at R, followed by an inverse-square decrease outside.

Worked example 3. A body weighs 250 N at the surface. Assuming uniform Earth density, find its weight halfway from the surface to the centre.

Formula: Wᵈ = W(1 − d/R), where Wᵈ denotes weight at depth.

Substitute: d = R/2, so Wᵈ = 250(1 − 1/2).

Answer: Wᵈ = 125 N. The smaller enclosed attracting mass must be included in the calculation.

Why does gravity vary with latitude?

Latitude is angular position north or south of the equator. Earth’s rotation reduces effective gravity, the acceleration inferred from weight on the rotating Earth. This reduction is greatest at the equator and zero at the poles.

Earth’s equatorial radius is also greater than its polar radius. The combined effect is that surface gravity is greater at the poles than at the equator. These latitude effects are separate from the spherical, non-rotating model used in the altitude and depth derivations.

What are gravitational potential and potential difference?

Gravitational potential V at a point is gravitational potential energy per unit mass. Equivalently, with zero potential at infinity, it is the external work per unit mass in bringing a test mass slowly from infinity to that point without changing its kinetic energy, the energy of motion.

The SI unit of gravitational potential is J kg⁻¹, where J denotes the joule, the unit of energy or work. One joule is one newton metre: 1 J = 1 N m. Potential is a scalar, a quantity having magnitude and sign without a spatial direction.

For points A and B outside Earth, let rₐ and rᵦ be their respective distances from its centre. Let Vₐ and Vᵦ denote their potentials. The change in potential is written ΔV = Vᵦ − Vₐ; Δ means final value minus initial value.

Derivation: Potential difference by integration

1. Take outward displacement as positive. At radial distance r, gravity acts inward, so a slowly applied external force per unit mass has outward magnitude GM/r².
2. For an outward displacement dr, where dr is an infinitesimally small change in r, the external work per unit mass is (GM/r²)dr.
3. Integrate from rₐ to rᵦ: Vᵦ − Vₐ = ∫ from rₐ to rᵦ (GM/r²)dr. The symbol ∫ denotes integration, which sums the small work contributions.
4. The integral of r⁻² with respect to r is −1/r. Evaluating at the limits gives GM(1/rₐ − 1/rᵦ).

ΔV = GM(1/rₐ − 1/rᵦ)

Set the initial point at infinity, written ∞, and assign its potential the value zero. Since 1/rₐ tends to zero there, the potential at external radius r becomes V(r) = −GM/r.

What does the negative sign mean?

With the zero at infinity, potential is negative at a finite distance from the attracting mass. Gravity does positive work as a mass moves inward, while the slow external agent does negative work. Moving outward requires positive external work and raises the potential towards zero.

The potential increases with distance even though its magnitude decreases. This distinction prevents a common sign error: a less negative value is a larger potential. The zero is a convention; changing it does not change the gravitational force or any potential difference.

How is gravitational potential energy related to height?

Gravitational potential energy U is energy associated with the positions of attracting masses. For a mass m in Earth’s field, U = mV = −GMm/r outside the spherical Earth, taking zero energy at infinity.

The SI unit of gravitational potential energy is J. Potential V describes energy per unit mass at a position; potential energy U also depends on the mass placed there. Doubling that mass doubles U without changing Earth’s potential at the point.

How does the familiar near-surface expression follow?

Consider lifting mass m from the surface to height h. The energy change is ΔU = GMm[1/R − 1/(R + h)]. Combining the fractions gives ΔU = GMmh/[R(R + h)].

If h ≪ R, replace R + h by R in the denominator. Then ΔU ≈ GMmh/R². Using g = GM/R² gives ΔU ≈ mgh. This is a near-surface energy difference, not the general potential energy measured from infinity.

The approximation treats gravity as practically constant during the displacement. For larger distances, integrate the varying force or use the exact difference of the inverse-distance expressions. A single surface value of g no longer describes the entire journey.

How does work relate to energy?

Gravity is a conservative force: its work between two positions is independent of the path. Work done by gravity equals the negative of the change in potential energy. For slow lifting with unchanged kinetic energy, the external work equals the positive change in potential energy.

Note: A zero gravitational field does not necessarily mean zero potential. Fields add as vectors, while potentials add as scalars. At a symmetric point between equal masses, their force contributions can cancel while their negative potential contributions add.

For a system containing several particles, total gravitational potential energy is the sum over all distinct pairs. Count each pair once. Potential at a specified point is instead found by adding the potentials due to each source mass at that point.

How is escape speed derived from conservation of energy?

Escape speed vₑ is the minimum initial speed needed for a body to reach infinity without further propulsion. The term escape velocity is often used, but the result specifies a speed rather than a particular velocity vector.

Use an isolated spherical Earth model, neglect air resistance and other celestial bodies, and ignore Earth’s rotation in the surface calculation. Mechanical energy E is the sum of kinetic energy K, the energy of motion, and gravitational potential energy U.

Derivation: Escape from Earth’s surface

1. For launch speed vₑ, the kinetic energy at the surface is K = ½mvₑ². Potential energy there is U = −GMm/R.
2. For the limiting escape, the speed tends to zero at infinity, where potential energy is chosen as zero. The final mechanical energy is therefore zero.
3. Conservation of mechanical energy gives ½mvₑ² − GMm/R = 0. Rearrange and cancel the projectile mass m.

vₑ = √(2GM/R) = √(2gR)

The symbol √ denotes the positive square root. Escape speed is independent of the projectile’s mass in this model. For a launch at external radius r, the corresponding expression is √(2GM/r), so launch position matters.

Worked example 4. Calculate the surface escape speed using g = 9.8 m s⁻² and Earth’s radius R = 6.4 × 10⁶ m. Neglect air resistance and Earth’s rotation.

Formula: vₑ = √(2gR).

Substitute: vₑ = √(2 × 9.8 × 6.4 × 10⁶).

Answer: vₑ ≈ 1.12 × 10⁴ m s⁻¹ = 11.2 km s⁻¹, where km denotes kilometre.

Why is escape easier from the Moon?

The Moon’s surface gravity and radius are both smaller than Earth’s. Its escape speed is about 2.3 km s⁻¹, about five times smaller than Earth’s. Gas molecules formed at its surface with speeds exceeding this value can escape its gravitational pull.

This explains the absence of an atmosphere on the Moon in this treatment. Escape speed depends on the ratio of the source’s mass to its radius, so comparisons between arbitrary bodies must consider both quantities, rather than mass alone.

How are a satellite’s orbital speed and period calculated?

A satellite is a body orbiting another body. The Moon is a natural Earth satellite; an artificial satellite is placed in orbit by human action. For the following calculation, assume a circular orbit and a satellite mass much smaller than Earth’s mass.

Let vₒ denote orbital speed and r = R + h the orbital radius. The required centripetal acceleration, the inward acceleration that changes the direction of motion, is vₒ²/r. Gravity supplies it.

Derivation: Circular orbital speed and period

1. For satellite mass m, equate gravitational force to the required centripetal force: GMm/r² = mvₒ²/r.
2. Cancel m and one factor of r to obtain vₒ² = GM/r. Thus vₒ = √(GM/r).
3. The satellite travels one circumference, 2πr, per revolution. Here π is the ratio of a circle’s circumference to its diameter. Hence T = 2πr/vₒ.
4. Substitute the speed to obtain T = 2π√(r³/GM). Squaring gives T² = 4π²r³/(GM), consistent with Kepler’s period law.

vₒ = √[GM/(R + h)]; T = 2π√[(R + h)³/GM]

Higher circular orbits have lower speeds and longer periods. Neither expression depends on satellite mass. The satellite’s acceleration equals the gravitational acceleration at its altitude, rather than the surface value.

Worked example 5. Estimate speed and period for a circular orbit very close to Earth’s surface. Use g = 9.8 m s⁻² and R = 6.4 × 10⁶ m, and neglect atmospheric effects.

Formula: vₒ = √(gR); T = 2πR/vₒ.

Substitute: vₒ = √(9.8 × 6.4 × 10⁶) ≈ 7.92 × 10³ m s⁻¹; T = 2π × 6.4 × 10⁶/(7.92 × 10³).

Answer: vₒ ≈ 7.92 km s⁻¹ and T ≈ 5.08 × 10³ s, or approximately 85 minutes. This is an ideal near-surface estimate.

Worked example 6. Infer Earth’s mass from the Moon’s orbit, treated as circular. Use orbital radius r = 3.84 × 10⁸ m, period 27.3 days, G = 6.67 × 10⁻¹¹ N m² kg⁻² and π = 3.14. One day is 86,400 s.

Formula: T = 27.3 × 86,400 s; M = 4π²r³/(GT²).

Substitute: M = 4 × 3.14² × (3.84 × 10⁸)³/[6.67 × 10⁻¹¹ × (27.3 × 86,400)²].

Answer: The converted period is 2,358,720 s, giving M ≈ 6.02 × 10²⁴ kg. This is close to the estimate obtained from surface gravity.

Why does an orbiting astronaut feel weightless?

Weightlessness in an orbiting spacecraft means absence of the usual supporting force on the astronaut. It does not mean that Earth’s gravitational field has vanished. The astronaut and spacecraft are both falling freely towards Earth while moving around it.

Free fall is motion under gravity without a supporting force. The normal reaction is the contact force exerted by a supporting surface. In ideal orbital free fall, the astronaut does not require a floor to provide support against gravity.

A weighing scale measures this support force, often called apparent weight. Its reading can be zero while the gravitational force remains non-zero. Gravity supplies the inward acceleration necessary for the orbital motion of both the spacecraft and its contents.

How does the satellite’s energy show that it is bound?

A bound orbit keeps the satellite at finite distances from Earth. For a circular orbit, substituting vₒ² = GM/r into K = ½mvₒ² gives K = GMm/(2r). The potential energy is U = −GMm/r.

Adding them gives E = K + U = −GMm/(2r). The kinetic energy is positive; the potential energy is negative and has twice its magnitude. The total mechanical energy is negative when the zero of potential energy is at infinity.

To escape, the satellite must gain enough energy for the total to reach zero. A satellite in a higher circular orbit has a less negative total energy, even though its orbital speed and kinetic energy are smaller.

What changes in an elliptical orbit?

In an elliptical orbit, distance and speed vary, so kinetic and potential energies each vary. Their sum remains constant when gravity is the only force doing work. Conservation of total energy does not mean that each part stays separately constant.

Keep these ideas distinct: zero apparent weight describes the support force, while negative mechanical energy describes a bound gravitational orbit. Neither statement implies zero gravitational acceleration.

What distinguishes geostationary and polar satellites?

A geostationary satellite appears fixed above the same point on Earth’s equator. It revolves with Earth’s rotation, so a ground observer sees an unchanged direction to it. Matching the rotation period is necessary, but is not the only condition.

What conditions give a geostationary orbit?

• The orbit must be circular, with its centre at Earth’s centre.
• Its orbital plane must coincide with Earth’s equatorial plane.
• It must revolve in the same direction as Earth rotates, from west to east.
• Its orbital period must equal Earth’s rotation period, approximately 24 hours for the following estimate.

The parking orbit here means the geostationary orbit in which the satellite remains above a fixed equatorial location. Its radius follows from the circular-orbit period equation. The radius is measured from Earth’s centre; the height is measured from the surface.

Worked example 7. Estimate geostationary orbital radius and height. Use T = 24 hours = 86,400 s, g = 9.8 m s⁻² and R = 6.4 × 10⁶ m, with a spherical Earth.

Formula: r = [gR²T²/(4π²)]⅓; h = r − R. The exponent ⅓ denotes a cube root.

Substitute: r = [9.8 × (6.4 × 10⁶)² × 86,400²/(4π²)]⅓ ≈ 4.23 × 10⁷ m.

Answer: r ≈ 42,300 km and h ≈ 35,900 km. These are approximate values based on the stated rounded inputs and 24-hour period.

How does a polar satellite differ?

A polar satellite has an orbit passing near the poles, with its orbital plane approximately perpendicular to the equatorial plane. As Earth rotates beneath it, successive passes view different strips of the surface.

Polar satellites are useful for remote sensing, meaning collecting information about Earth from a distance, including surface mapping and weather observations. Geostationary satellites are useful for communications because they maintain a fixed apparent position relative to the ground.

Glossary

• Gravitation — The mutual attractive interaction between masses, described for point masses by Newton’s inverse-square law.
• Gravitational field intensity — Gravitational force per unit test mass at a point, directed along the gravitational acceleration.
• Weight — The gravitational force on a body, equal to its mass multiplied by local gravitational acceleration.
• Ellipse — A closed curve whose points have a constant sum of distances from two fixed foci.
• Semi-major axis — Half the longest diameter of an ellipse, used to describe the size of an elliptical orbit.
• Central force — A force acting along the line joining a moving body to a fixed centre.
• Gravitational potential — Gravitational potential energy per unit mass at a point, with a specified choice of zero.
• Conservative force — A force whose work between two positions is independent of the path followed.
• Escape speed — Minimum initial speed required to reach infinity without further propulsion under the stated gravitational assumptions.
• Orbital period — Time required for a satellite or planet to complete one full revolution around its central body.
• Weightlessness — Absence of apparent weight when a body has no supporting force, as in ideal orbital free fall.
• Geostationary satellite — A satellite in an equatorial circular orbit that remains above the same point on the rotating Earth.
• Polar satellite — A satellite passing near Earth’s poles and viewing different surface regions as Earth rotates beneath it.

Common errors and misconceptions

• Misconception: G and g are the same quantity. Correct: G is the universal gravitational constant; g is gravitational acceleration at a particular position. Their units differ.
• Misconception: Altitude is the distance used in the inverse-square formula. Correct: For Earth, use distance from the centre, R + h, rather than height h above the surface.
• Misconception: Gravity increases below the surface because the centre is nearer. Correct: For uniform density it decreases with depth because only the enclosed mass contributes to the net force.
• Misconception: Gravitational potential is positive because energy is required to escape. Correct: With zero at infinity, potential is negative; positive external work raises it towards zero.
• Misconception: The expression mgh applies at any height. Correct: It approximates a potential-energy difference near the surface, where gravity can be treated as practically constant.
• Misconception: Astronauts are weightless because gravity is absent. Correct: Gravity accelerates both spacecraft and astronaut in free fall; the supporting force and apparent weight vanish.
• Misconception: Any satellite with a 24-hour period is geostationary. Correct: It must also follow a circular equatorial orbit in Earth’s direction of rotation.

Exam-style questions with model answers

Q1. State Newton’s universal law of gravitation and give the SI unit and dimensional formula of G, the universal gravitational constant. [2 marks]

1. Two point masses attract with force proportional to their mass product and inversely proportional to their separation squared, directed along their joining line.
2. The SI unit of G is N m² kg⁻² and its dimensional formula is [M⁻¹L³T⁻²], where M, L and T denote mass, length and time dimensions.

Q2. State Kepler’s three laws of planetary motion, defining the orbital quantities used. [3 marks]

1. Planets move in ellipses with the Sun at one focus. An ellipse has a constant sum of distances from two fixed points called foci.
2. The Sun-planet line sweeps equal areas in equal time intervals. Thus the planet moves faster near the Sun and slower farther away.
3. The square of the time for one revolution is proportional to the cube of the semi-major axis, which is half the ellipse’s longest diameter.

Q3. A body weighs 63 N on the surface of a spherical, non-rotating Earth of radius R. Find its gravitational force at height h = R/2, using the inverse-square law. [3 marks]

1. The new distance from Earth’s centre is R + h = 3R/2. The inverse-square law requires this centre-to-body distance rather than the height above the surface.
2. With unchanged body mass, the ratio of new force to surface weight is [R/(3R/2)]² = 4/9. Earth’s mass and the gravitational constant cancel from the ratio.
3. The new force is 63 × 4/9 = 28 N, directed towards Earth’s centre. The decrease is due to the weaker gravitational field at the greater distance.

Q4. Derive gravitational acceleration at depth d in a spherical Earth of uniform density, mass M and radius R. Let G be the universal gravitational constant and g = GM/R² the surface acceleration; neglect rotation. Explain the central value. [5 marks]

1. A point at depth d is at radius r = R − d from the centre. Divide Earth into the sphere inside this radius and the surrounding spherical shells.
2. The surrounding shells give zero net gravitational force at the point. Therefore only the enclosed sphere contributes to the local gravitational acceleration.
3. Let Mᵣ be its mass. Uniform density makes the mass ratio equal to the volume ratio, so Mᵣ/M = r³/R³ and Mᵣ = Mr³/R³.
4. The acceleration magnitude is GMᵣ/r² = GMr/R³. Substituting the given surface relation gives gᵈ = gr/R = g(1 − d/R), where gᵈ denotes acceleration at depth.
5. At the centre d = R, so the acceleration is zero. The linear decrease with depth depends on the assumed uniform density and should not be presented as an exact real-Earth profile.

Q5. For a spherical Earth of mass M and radius R, derive the potential difference between external points A and B at radii rₐ and rᵦ. Use the universal gravitational constant G, zero potential at infinity and slow external displacement without a kinetic-energy change. [5 marks]

1. Gravitational potential is external work per unit mass in bringing a test mass from infinity to the point under the stated slow-displacement condition. Let the potentials at A and B be Vₐ and Vᵦ.
2. At radius r, external force per unit mass balancing gravity has outward magnitude GM/r². For an outward infinitesimal displacement dr, the work per unit mass is (GM/r²)dr.
3. Integrate along the radial path from rₐ to rᵦ: Vᵦ − Vₐ = ∫ from rₐ to rᵦ (GM/r²)dr. Gravity’s conservative character makes the result independent of path.
4. Integration gives Vᵦ − Vₐ = GM(1/rₐ − 1/rᵦ). Moving to a larger radius therefore raises potential, because the expression is positive.
5. Putting the initial point at infinity and its potential at zero gives V(r) = −GM/r. The negative value expresses attractive gravity with this chosen reference.

Q6. Derive the surface escape speed from a spherical, non-rotating Earth of mass M and radius R. Let G be the universal gravitational constant and m the projectile mass. Neglect air resistance, other bodies and further propulsion, and take potential energy as zero at infinity. [5 marks]

1. Escape speed vₑ is the minimum initial speed for the projectile to reach infinity. For this limiting case, its speed tends to zero there.
2. The initial kinetic energy is ½mvₑ² and the initial gravitational potential energy is −GMm/R. Their sum is the initial mechanical energy.
3. At infinity, both kinetic and potential energies are zero in the limiting case. With no dissipative work or further propulsion, mechanical energy is conserved.
4. Equate the energies: ½mvₑ² − GMm/R = 0. Cancelling m and rearranging gives vₑ² = 2GM/R, and therefore vₑ = √(2GM/R).
5. The projectile mass cancels, so the required speed is independent of m under these assumptions. It depends on the source mass and launch radius, rather than on projectile mass.

Q7. A satellite moves in a circular orbit of radius r about a spherical Earth of mass M, with satellite mass negligible compared with M. Derive its speed and period using G, the universal gravitational constant, and explain why an astronaut aboard feels weightless. [4 marks]

1. For satellite mass m and orbital speed vₒ, gravity provides the centripetal force: GMm/r² = mvₒ²/r.
2. Cancelling m and rearranging gives vₒ = √(GM/r). The circular orbital speed is independent of satellite mass under the stated approximation.
3. One orbit has length 2πr, where π is the circle constant. The period is T = 2πr/vₒ = 2π√(r³/GM).
4. The astronaut and spacecraft fall freely together. Their gravitational acceleration is non-zero, but the astronaut needs no supporting force from the floor and therefore has zero apparent weight.

Q8. State four conditions that an Earth satellite must satisfy to be geostationary. [4 marks]

1. Its orbit must be circular about Earth’s centre. A changing orbital radius would prevent the required constant angular motion relative to Earth.
2. Its orbital plane must be Earth’s equatorial plane, so it stays above the equator rather than moving north and south.
3. It must move in the same direction as Earth’s rotation, from west to east.
4. Its orbital period must equal Earth’s rotation period. Together these conditions keep it above the same equatorial point.

Key takeaways

• Universal gravitation is an attractive inverse-square interaction; centre-to-centre distance is used for external points around a spherical source.
• Kepler’s laws connect elliptical orbits, equal swept areas in equal times, and the relation between orbital period and semi-major axis.
• Gravitational field intensity equals gravitational acceleration; weight depends on local gravity, while the body’s mass does not change.
• Gravity decreases with altitude, and decreases linearly with depth only under the uniform-density spherical-Earth assumption.
• Gravitational potential and potential energy are negative when their reference value at infinity is chosen to be zero.
• The familiar mgh expression approximates a near-surface potential-energy change and requires height to be much smaller than Earth’s radius.
• Escape speed follows from energy conservation, while circular orbital speed follows by equating gravitational force with required centripetal force.
• Orbital weightlessness results from shared free fall; a geostationary satellite additionally requires a circular equatorial orbit matching Earth’s rotation.

Test yourself

Why must height be added to Earth’s radius in an external gravity calculation?

The inverse-square law uses distance from Earth’s centre. Height above the surface alone omits Earth’s radius.

What assumption is needed for a linear decrease of gravity with depth?

Earth must be treated as a sphere of uniform density, with rotation neglected in this model.

Why does an outer spherical shell not contribute to the depth formula?

A uniform spherical shell produces zero resultant gravitational force at a point anywhere inside it.

How can potential increase as its magnitude decreases?

Potential is negative with zero at infinity. A value closer to zero is larger even though its absolute magnitude is smaller.

Does escape speed depend on the mass of the escaping object?

No. The object’s mass cancels from the energy equation under the ideal gravitational assumptions.

What happens to speed and period in a higher circular orbit?

Orbital speed decreases and orbital period increases for satellites orbiting the same central mass.

Why is an astronaut’s zero scale reading not evidence of zero gravity?

The scale measures support force. Astronaut and spacecraft fall together while gravity supplies their orbital acceleration.

Why is matching Earth’s rotation period insufficient for geostationarity?

The orbit must also be circular, equatorial and in the same direction as Earth’s rotation.