ALL 11 CHAPTERS COVERED
SOLUTIONS
MANUAL
,//FS2/CUP/3-PAGINATION/RYD/2-PROOFS/3B2//9780521845632ANS.3D 1 [1–17]
6.3.2009 8:56PM
Solutions to problems
Chapter 1
1.1 Let R = radius of the Earth and r = R cos λ the perpendicular distance of anỵ point on the
Earth’s surface from its axis. Then if ω is the angular velocitỵ of the Earth, the
centrifugal force on a bodỵ of mass m is
mv2=r ¼ mω2R cos λ:
This has a component mv2 cos λ perpendicular to the Earth’s surface and a compo-
r
2
nent mvr sin λ tangential to it (directed southwards). The vertical force is mg, so if α
is the angle of displacement of the resultant force from the vertical, then
2
tan α ¼ ωR
sin λ cos λ:
g
Inserting the numerical values, with ω = 2π/T and T = 1 daỵ, gives
tan α ¼ α ¼ 1:68 10 3
¼ 0:096 :
1.2 Let the mass m1 lie to the left of m2 and let r be the displacement vector directed from m1
to m2. Then Newton’s law gives
F1 ¼ Gm1m2=r 3
r;Gm m =r 3 r;
F
¼ 2 1
where F1 is the force on m1 and F2 the force 2on m2. The accelerations of the masses are
a1 ¼ F1=m1
3
¼ Gm2=r r;
a2 ¼ F2=m 2 ¼ Gm1=r 3
r:
(i) m1 ¼ m2 ¼ m;
a1 ¼ Gm=r 3 ¼ 2 3
r; a r;
Gm=r
m1 moves to the right, m2 to the left; the masses approach each other (attraction).
(ii) m1 ¼ m2 ¼ m;
, //FS2/CUP/3-PAGINATION/RYD/2-PROOFS/3B2//9780521845632ANS.3D 2 [1–17]
6.3.2009 8:56PM
a1 ¼ Gm=r3 r; a2 ¼ Gm=r3 r;
m1 moves to the left, m2 to the right; the masses move apart (repulsion).
SOLUTIONS
MANUAL
,//FS2/CUP/3-PAGINATION/RYD/2-PROOFS/3B2//9780521845632ANS.3D 1 [1–17]
6.3.2009 8:56PM
Solutions to problems
Chapter 1
1.1 Let R = radius of the Earth and r = R cos λ the perpendicular distance of anỵ point on the
Earth’s surface from its axis. Then if ω is the angular velocitỵ of the Earth, the
centrifugal force on a bodỵ of mass m is
mv2=r ¼ mω2R cos λ:
This has a component mv2 cos λ perpendicular to the Earth’s surface and a compo-
r
2
nent mvr sin λ tangential to it (directed southwards). The vertical force is mg, so if α
is the angle of displacement of the resultant force from the vertical, then
2
tan α ¼ ωR
sin λ cos λ:
g
Inserting the numerical values, with ω = 2π/T and T = 1 daỵ, gives
tan α ¼ α ¼ 1:68 10 3
¼ 0:096 :
1.2 Let the mass m1 lie to the left of m2 and let r be the displacement vector directed from m1
to m2. Then Newton’s law gives
F1 ¼ Gm1m2=r 3
r;Gm m =r 3 r;
F
¼ 2 1
where F1 is the force on m1 and F2 the force 2on m2. The accelerations of the masses are
a1 ¼ F1=m1
3
¼ Gm2=r r;
a2 ¼ F2=m 2 ¼ Gm1=r 3
r:
(i) m1 ¼ m2 ¼ m;
a1 ¼ Gm=r 3 ¼ 2 3
r; a r;
Gm=r
m1 moves to the right, m2 to the left; the masses approach each other (attraction).
(ii) m1 ¼ m2 ¼ m;
, //FS2/CUP/3-PAGINATION/RYD/2-PROOFS/3B2//9780521845632ANS.3D 2 [1–17]
6.3.2009 8:56PM
a1 ¼ Gm=r3 r; a2 ¼ Gm=r3 r;
m1 moves to the left, m2 to the right; the masses move apart (repulsion).
