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Every pair of masses attracts each other along the line joining their centres. The force is proportional to the product of the two masses and inversely proportional to the square of the distance between their centres. The two bodies pull on each other with equal and opposite forces.
G is the universal gravitational constant, 6.67 × 10−11 N m2 kg−2, and it has the same value everywhere. Its dimensional formula is [M−1L3T−2].
Show derivationObservation gives two facts: F ∝ m1m2 and F ∝ 1/d2. Combining them, F ∝ m1m2/d2. Replacing the proportionality sign with a constant, which we call G, gives F = G m1m2/d2.
The acceleration of a freely falling body near the Earth's surface is called the acceleration due to gravity, g. For the Earth, with mass M and radius R:
g does not depend on the mass of the falling body, so a feather and a stone fall together when there is no air resistance. Putting in M = 5.97 × 1024 kg and R = 6.37 × 106 m gives g ≈ 9.8 m s−2.
Show derivationAt the surface the pull on a body of mass m is GMm/R2. By Newton's second law the same force is mg. Equating them, mg = GMm/R2. The m cancels, leaving g = GM/R2.
The familiar 9.8 m s−2 is only an average. The value of g changes with height, depth and position on the Earth.
g gets smaller as you go up. This form is accurate only when h is much smaller than R. For larger heights use g′ = gR2/(R + h)2.
Show derivationAt height h the distance from the centre is R + h, so g′ = GM/(R + h)2. Dividing by g = GM/R2 gives g′/g = R2/(R + h)2 = (1 + h/R)−2. For small h/R the binomial expansion gives (1 + h/R)−2 ≈ 1 − 2h/R, so g′ = g(1 − 2h/R).
Going down a mine shaft, g decreases and becomes zero at the centre of the Earth (x = R). This assumes the Earth has uniform density.
Show derivationWriting M = (4/3)πR3ρ, the surface value is g = (4/3)πGρR. At depth x only the inner sphere of radius (R − x) pulls the body, because the shell above it has no net effect. So g′ = (4/3)πGρ(R − x). Dividing, g′/g = (R − x)/R = 1 − x/R.
The Earth bulges at the equator, so its polar radius is smaller than its equatorial radius. Since g ∝ 1/R2, g is larger at the poles than at the equator (gp > ge). The Earth's spin lowers the effective value at the equator a little more.
The region around a mass where another mass feels a gravitational force is its gravitational field. The field intensity at a point is the force on a unit test mass placed there, and it points towards the mass producing the field.
On the Earth's surface r = R, so E = GM/R2 = g. Field intensity is measured in N kg−1, which is the same as m s−2, so on the surface it equals the acceleration due to gravity.
Show derivationA test mass m at distance r from the centre of the Earth feels F = GMm/r2. Dividing by m gives E = GM/r2.
The gravitational potential at a point is the work done by an external agent in bringing a unit mass from infinity to that point without changing its kinetic energy. It is zero at infinity and negative everywhere else, because gravity is attractive.
At distance x from the centre, gravity pulls a unit mass inwards with force GM/x2. The agent must pull outwards with the same force. Moving the unit mass from x = ∞ in to x = r, the work done by the agent is W = GM × [−1/x] evaluated from ∞ to r, which equals −GM/r. This work is V.
The gravitational potential energy of a mass m at a point is the work done in bringing it there from infinity. It equals m times the potential at that point.
Close to the ground, lifting a mass from R up to R + h raises its potential energy by GMm(1/R − 1/(R + h)), which is approximately mgh when h is small.
Show derivationRepeat the derivation for potential with a mass m: the force becomes GMm/x2 and the work from infinity to r is −GMm/r. For small h, GMm(1/R − 1/(R + h)) = GMm·h / (R(R + h)) ≈ GMm·h / R2 = mgh.
The escape velocity is the minimum speed with which a body must be launched from the Earth's surface to get beyond the Earth's gravitational pull and never fall back.
With g = 9.8 m s−2 and R = 6.4 × 106 m, ve ≈ 11.2 km s−1. It does not depend on the mass of the body.
Show derivationTo just reach infinity with no speed left, the starting kinetic energy must cancel the potential energy: ½mv2 − GMm/R = 0. So v2 = 2GM/R. Using GM = gR2, v = √(2gR).
A satellite is a body that revolves around a planet along a closed path called its orbit. Its speed along the orbit is the orbital velocity. For a circular orbit of radius r = R + h:
Just above the surface (r ≈ R) this is √(gR) ≈ 7.9 km s−1. The orbital speed does not depend on the satellite's own mass, and it gets smaller for higher orbits.
Show derivationGravity supplies the centripetal force: GMm/r2 = mv02/r. Cancelling m and one r gives v02 = GM/r. With GM = gR2, v0 = R√(g/r).
The time period T is the time taken for one complete revolution.
So T2 ∝ r3, which is Kepler's third law. Close to the ground the period is about 84 minutes. To find the height of a satellite whose period is known:
From T = 2πr/v0 and v0 = R√(g/r) we get T = (2πr/R)√(r/g), so T2 = 4π2r3/(gR2). Hence r3 = gR2T2/4π2. Since r = R + h, h = r − R.
A satellite has kinetic energy because it moves, and potential energy because of its position in the Earth's field.
The total energy is negative, so the satellite is bound to the Earth. Its size, GMm/2r, is the extra energy needed to free the satellite from its orbit.
Show derivationFrom GMm/r2 = mv2/r we get mv2 = GMm/r, so K.E. = ½mv2 = GMm/2r. The potential energy is −GMm/r. Adding them, E = GMm/2r − GMm/r = −GMm/2r.
A geostationary satellite appears fixed at one point in the sky when seen from the Earth. Its orbit is called a parking orbit, and it must meet three conditions:
Its height is about 36,000 km above the surface (roughly 42,000 km from the Earth's centre) and its speed is about 3.1 km s−1. Communication and weather satellites use this orbit.
Show workingPut T = 24 h = 86 400 s into the height formula, with g = 9.8 m s−2 and R = 6.4 × 106 m. Then r3 = gR2T2/4π2 ≈ 7.6 × 1022 m3, so r ≈ 4.23 × 107 m. The height is h = r − R ≈ 3.6 × 107 m, about 36,000 km.
A planet has 8 times the mass of the Earth and 2 times its radius. Find g on the planet and its escape velocity. (On the Earth, g = 9.8 m s−2 and ve = 11.2 km s−1.)
Show solutiong ∝ M/R2, so gp = 9.8 × 8/22 = 19.6 m s−2.
ve ∝ √(M/R), so vp = 11.2 × √(8/2) = 22.4 km s−1.
At what height above the Earth's surface does g drop to one quarter of its surface value?
Show solutionUse the exact relation: g′/g = R2/(R + h)2 = 1/4, so (R + h)/R = 2 and h = R ≈ 6400 km.
The approximation 1 − 2h/R would give the wrong answer here, because h is not small compared with R.
A satellite orbits 300 km above the Earth. Find its orbital speed and period. Take R = 6400 km and g = 9.8 m s−2.
Show solutionr = R + h = 6.7 × 106 m.
v0 = R√(g/r) = 6.4 × 106 × √(9.8 / 6.7 × 106) ≈ 7.74 × 103 m s−1.
T = 2πr / v0 ≈ 5.4 × 103 s, which is about 91 minutes.
A body weighs 600 N on the Earth's surface. What does it weigh at a depth of R/4 below the surface?
Show solutionWeight is proportional to g′ = g(1 − x/R) = g(1 − 1/4) = 3g/4.
W′ = 600 × 3/4 = 450 N.