The effective value of acceleration due to gravity g' = g cos \(\alpha\)
\(T=\quad 2\pi \sqrt { \frac { l }{ g' } } =2\pi \sqrt { \frac { 1 }{ cos\alpha } } \)

Mass of the earth M = 6.0\(\times\)10²⁴ kg ; m = 67 kg G = 6.67\(\times\)10^-11 Nm² kg-2
GravitationaI potentia V = -\(\frac{GM}{R}\)
\(=\frac { 6.67\times { 10 }^{ -11 }\times 6\times { 10 }^{ 24 } }{ 6.6\times { 10 }^{ 10 } } \)
V = -6.1\(\times\)103 Jkg^-1

(i) The seasons in the Earth arise due to the rotation of Earth around the Sun with 23.5° tilt.
(ii) Due to this 23.5° tilt, when the northern part of Earth is farther to the Sun, the southern part is nearer to the Sun.
(iii) So when it is summer in the northern hemisphere, the southern hemisphere experience winter.

A satellite of mass M to move in a circular orbit, centripetal force must be acting on the satellite. This centripetal force is provided by the Earth's gravitational force.
\(\frac { { mv }^{ 2 } }{ \left( { R }_{ E }+h \right) } =\frac { { GMM }_{ E } }{ { \left( { R }_{ e }+h \right) }^{ 2 } } \)
\({ v }^{ 2 }=\frac { { GM }_{ E } }{ \left( { R }_{ E }+h \right) } \)
\(v=\sqrt { \frac { { GM }_{ E } }{ \left( { R }_{ E }+h \right) } } \)
As h increases, the speed of the satellite decreases.

(i) Consider two masses m₁ and m₂ separated by a distance r which has gravitational potential energy U(r).
(ii) The gravitational potential due to mass ml at a point P which is at a distance r from ml is obtained by making m₂ equal to unity (m₂=1kg).
(iii) Thus the gravitational potential V(r) due to mass m = 1 at a distance r is,
V(r)=\(\frac { { GM }_{ 1 } }{ { r } } \)