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Physics · Ch 5 — Gravitation

Acceleration due to Gravity

5.5

Acceleration due to Gravity

Section 5.3 gave the gravitational force between two point masses as F=GmMr2F=\dfrac{GmM}{r^2}. Since the Earth is, to a very good approximation, a uniform sphere, its entire mass M can (by the shell theorem of section 5.3) be treated as though concentrated at its centre when computing the force on an external point object of mass m at distance r from that centre: F=GMmr2F=\dfrac{GMm}{r^2}.

If no other force acts on the object, Newton's second law F=maF=ma lets this force be converted directly into an acceleration: a=Fm=GMr2a=\dfrac{F}{m}=\dfrac{GM}{r^2}. This acceleration -- towards the centre of the Earth -- is called the acceleration due to gravity, denoted g. Crucially, the object's OWN mass m cancelled out of the expression entirely: g=GMr2g=\dfrac{GM}{r^2} depends only on the Earth's mass M and the distance r, never on the mass of the falling object itself. This is exactly the property Galileo discovered experimentally: bodies of different mass, dropped from the same height, fall with identical acceleration.

When the object is close to the Earth's surface, r≈Rr\approx R (the Earth's radius), giving the familiar surface value gsurface=GMR2≈9.8g_{surface}=\dfrac{GM}{R^2}\approx9.8 m/s^2 -- treated as effectively constant for objects near the surface, since the small variation in their distance from the Earth's centre over ordinary heights (a building, a hill) is negligible compared to R itself. …

Misc Ex.4Example 5.4: Mass of the Earth calculated from g, R and G

Worked out. Given g=9.81g=9.81 m/s^2, RE=6.37×106R_E=6.37\times10^6 m and G=6.67×10−11G=6.67\times10^{-11} N m^2/kg^2, rearranging g=GME/RE2g=GM_E/R_E^2 to ME=gRE2/GM_E=gR_E^2/G gives ME=(9.81×(6.37×106)2)/(6.67×10−11)≈5.97×1024M_E=(9.81\times(6.37\times10^6)^2)/(6.67\times10^{-11})\approx5.97\times10^{24} kg, matching the accepted mass of the Earth -- demonstrating that the same law used for orbits and falling apples can be inverted to 'weigh' the entire planet …

Misc Ex.5Example 5.5: Acceleration due to gravity on the surface of the Moon

Worked out. Given the Moon's mass is 1/80 that of the Earth (Mm=M/80M_m=M/80) and its diameter (hence radius) is 1/4 that of the Earth (Rm=R/4R_m=R/4), the ratio gm/g=(Mm/M)×(R/Rm)2=(1/80)×16=16/80=1/5g_m/g=(M_m/M)\times(R/R_m)^2=(1/80)\times16=16/80=1/5, so gm=g/5=9.8/5=1.96g_m=g/5=9.8/5=1.96 m/s^2 -- the well-known result that lunar gravity is roughly one-sixth (here approximated as one-fifth) of Earth's. …

Misc Ex.6Example 5.6: Acceleration due to gravity on a planet 10 times as massive and 20 times the radius of the Earth

Worked out. Given Mp=10MEM_p=10M_E and Rp=20RER_p=20R_E, using gp/gE=(Mp/ME)×(RE/Rp)2=10×(1/20)2=10/400=1/40g_p/g_E=(M_p/M_E)\times(R_E/R_p)^2=10\times(1/20)^2=10/400=1/40, so gp=gE/40=9.8/40=0.245g_p=g_E/40=9.8/40=0.245 m/s^2 -- illustrating that a much more massive planet can still have very weak surface gravity if its radius is large enough, since g falls off with the SQUARE of radius but rises only linearly with mass. …