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What is the magnitude of gravitational force between earth & 1 kg object on the surface of earth?
We know gravitation force F=Easy Open in App Solution Verified by Toppr R 2 GMm = (6.4×10 6 ) 2 6.67×10 −11 ×6×10 24 ×1 =9.8N
Where M = mass of the earth, m = mass of the object and R = radius of earth
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What is the magnitude of the gravitational force between the earth and a 1 kg object on its surface? Mass of the earth is 6 × 1024 kg and radius of the earth is 6.4 × 106 m
What is the magnitude of the gravitational force between the earth and a 1 kg object on its surface? Mass of the earth is 6 × 1024 kg and radius of the earth is 6.4 × 106 m
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What is the magnitude of the gravitational force between the earth and a 1 kg object on its surface? Mass of the earth is 6 × 1024 kg and radius of the earth is 6.4 × 106 m
Question
Question 3What is the magnitude of the gravitational force between the earth and a 1 kg object on its surface? (Mass of the earth is
6 × 10 24 k g
and radius of the earth is
6.4 × 10 6 m ) Video Solution
NCERT - Grade 09 - Physics - Gravitation - Q3
PHYSICS
05:34 Min | 125 Views
Rate Open in App Solution Given that,
Mass of the body, m = 1 kg
Mass of the Earth, M = 6 × 10 24 k g
Radius of the Earth,
R = 6.4 × 10 6 m
From the universal law of gravitation, the magnitude of the gravitational force (F) between the earth and the body is given by,
F = G M × m R 2 where G = 6.67 × 10 − 11 N m 2 / k g 2
is the universal gravitational constant.
Hence, magnitude of the gravitational force (F) between the earth and the body is given by,
F = G M × m R 2 = 6.67 × 10 − 11 × 6 × 10 24 × 1 ( 6.4 × 10 6 ) 2
= 9.77 N = 9.8 N (approx).
Suggest Corrections 68
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स्रोत : byjus.com
What is the magnitude of the gravitational force between the earth and a 1 kg object on its surface ? (Mass of the earth is 6 xx 10^(24) kg and radius of the earth is 6.4 xx 10^(6) m)
Here, mass (m) of the body =1kg mass (Me) of the earth =6 xx 10^(24)kg, radius (R(e)) of the earth =6.4 xx 10^(6)m. Magnitude of the gravitional force between the earth and 1kg object. F=G ""(M(e)m)/(R(e)^(2))=(6.67xx 10^(-11) xx 6 xx 10^(24) xx1)/((6.4 xx 10^(6))^(2)) =(6.67 xx 10^(13))/(6.4 xx 6.4 xx 10^(12)) =(6.67 xx 6 xx 10)/(6.4 xx 6.4) =9.77N approx 9.8N
Home > English > Class 9 > Physics > Chapter > Gravitation >
What is the magnitude of the g...
What is the magnitude of the gravitational force between the earth and a 1 kg object on its surface ?
(Mass of the earth is
6× 10 24 6×1024
kg and radius of the earth is
6.4× 10 6 m) 6.4×106m)
Updated On: 27-06-2022
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Text Solution Open Answer in App Answer
The correct Answer is The magnitude of the gravitional force between the earth and a 1kg object is 9.8N.
Solution
Here, mass (m) of the body =1kg mass
( M e ) (Me) of the earth =6× 10 24 kg =6×1024kg , radius ( R e ) (Re) of the earth =6.4× 10 6 m =6.4×106m .
Magnitude of the gravitional force between the earth and 1kg object.
F=G M e m R 2 e = 6.67× 10 −11 ×6× 10 24 ×1 (6.4× 10 6 ) 2
F=GMemRe2=6.67×10-11×6×1024×1(6.4×106)2
= 6.67× 10 13 6.4×6.4× 10 12
=6.67×10136.4×6.4×1012
= 6.67×6×10 6.4×6.4 =9.77N≈9.8N
=6.67×6×106.4×6.4=9.77N≈9.8N
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