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Grade 12Mechanics

A body weighs 63N on the surface of the earth. The gravitational force on it due to the earth at a height equal to half of the radius of the
earth is 7xN, then x=---

suyash sunil rokade

8 Years agoGrade 12

4 Answers

Arun

Approved Tutor Answer8 Years ago

Weight of the body, W = 63 N
Acceleration due to gravity at height h from the Earth’s surface is given by the relation:
g‘ = g / [1 + ( h / R_e) ]²
Where,
g = Acceleration due to gravity on the Earth’s surface
Re = Radius of the Earth
For h = R_e / 2
g‘ = g / [(1 + (R_e / 2R_e) ]²
= g / [1 + (1/2) ]² = (4/9)g
Weight of a body of mass m at height h is given as:
W‘ = mg
= m × (4/9)g = (4/9)mg
= (4/9)W
= (4/9) × 63 = 28 N.

Hafiz shah s.r

Approved Tutor Answer8 Years ago

Weight on the surface of earth=63N

Mg=63N

On the surface of earth gravitational force = Mg=63N

M=63/g.

Gravitational force at a ht ,H =R/2=

Mg¹.

g¹=g{R²/(R+H)²}...on solving g¹=4g/9

:-Mg¹=M×4g/9

Where M=63/g.

=63/g×4g/9=7×4=28N

:-. X= 4.

Hope you understood....

Nabin

8 Years ago

Weight on earth surface=63N

Mass of the body=6.3(g=10)

height of the body=radius/2

now,

g1={r^2/(r+h)^2}g

=40/9

Then

Force=m*g1

7x=6.3*40/9

x=4

Abhinav Vaisoya

7 Years ago

We know that w=mg

therefore w=f

f=48n

Or

$48=\frac{GMm}{r^{2}}$

When distance from surface of the Earth is 6400,

r becomes doubled so,

new weight= $\frac{GMm}{2r^{2}}$

mg2= $\frac{GMm}{4r^{2}}$

mg2= $\frac{GMm}{r^{2}}*\frac{1}{4}$

mg2= $48*\frac{1}{4}$

mg2=12

So new weight will be 12N

Hope this helps you!

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