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JEE Main Physics Class 11 Gravitation Part 1 Questions
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© examsnet.com
Question : 59
Total: 100
Take the mean distance of the moon and the sun from the earth to be
0.4
×
10
6
k
m
and
150
×
10
6
k
m
respectively. Their masses are
8
×
10
22
k
g
and
2
×
10
30
k
g
respectively. The radius of the earth is
6400
k
m
. Let
Δ
F
1
be the difference in the forces exerted by the moon at the nearest and farthest points on the earth and
Δ
F
2
be the difference in the force exerted by the sun at the nearest and farthest points on the earth. Then, the number closest to
Δ
F
1
Δ
F
2
is:
[Online April 15, 2018]
2
6
10
–
2
0.6
Validate
Solution:
As we know, Gravitational force of attraction,
F
=
G
M
m
R
2
F
1
=
G
M
e
m
r
1
2
and
F
2
=
G
M
e
M
s
r
2
2
Δ
F
1
=
2
G
M
e
m
r
1
3
Δ
r
1
and
Δ
F
2
=
G
M
e
M
s
r
2
3
Δ
r
2
Δ
F
1
Δ
F
2
=
m
Δ
r
1
r
1
3
r
2
3
M
s
Δ
r
2
=
(
m
M
s
)
(
r
2
3
r
1
3
)
(
Δ
r
1
Δ
r
2
)
Using
Δ
r
1
=
Δ
r
2
=
2
R
earth
;
m
=
8
×
10
22
k
g
M
s
=
2
×
10
30
k
g
r
1
=
0.4
×
10
6
k
m
and
r
2
=
150
×
10
6
k
m
Δ
F
1
Δ
F
2
=
(
8
×
10
22
2
×
10
30
)
(
150
×
10
6
0.4
×
10
6
)
3
×
1
≅
2
© examsnet.com
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