Problem 1 : At what height from the surface of Earth will the acceleration due to gravity is reduced by 36 % from the value at the surface. Take, R = 6400 km.

Solution : The acceleration due to gravity decreases as we go vertically up from the surface. The reduction of acceleration by 36 % means that the height involved is significant. As such, we can not use the approximated expression of the effective accelerations for h<< R as given by :

g
′
=
g
1
−
2
h
R
g
′
=
g
1
−
2
h
R

Instead, we should use the relation,

⇒
g
′
=
g
1
+
h
R
2
⇒
g
′
=
g
1
+
h
R
2

Note that we have considered reference gravitational acceleration equal to acceleration on the surface. Now, it is given that :

⇒
g
′
=
0.64
g
⇒
g
′
=
0.64
g

Hence,

⇒
0.64
g
=
g
1
+
h
R
2
⇒
0.64
g
=
g
1
+
h
R
2

⇒
1
+
h
R
2
X
0.64
=
1
⇒
1
+
h
R
2
X
0.64
=
1

⇒
1
+
h
R
=
10
8
=
5
4
⇒
1
+
h
R
=
10
8
=
5
4

⇒
h
R
=
5
4
−
1
=
1
4
⇒
h
R
=
5
4
−
1
=
1
4

⇒
h
=
R
4
=
6400
4
=
1600
k
m
⇒
h
=
R
4
=
6400
4
=
1600
k
m

Note : If we calculate, considering h << R, then

⇒
0.64
g
=
g
1
−
2
h
R
⇒
0.64
g
=
g
1
−
2
h
R

⇒
0.64
R
=
R
−
2
h
⇒
0.64
R
=
R
−
2
h

⇒
h
=
R
1
−
0.64
2
=
0.18
R
=
0.18
X
6400
=
1152
k
m
⇒
h
=
R
1
−
0.64
2
=
0.18
R
=
0.18
X
6400
=
1152
k
m