write equations of motion in terms of acceleration due to gravity and solve 10 numerical examples about equation
Answers
Answer:
v=u+at
s=1/2at2
v2=u2+2as
Answer:
Acceleration due to Gravity Formula
We have no doubt seen gravity work in our life. After all, it is the force that is helping us to keep our feet on the ground. If we throw a ball in the upward direction in the air. Then it will come down on its own. Why? When the ball is going in upwards direction, its speed will be less as compared to when it comes down. This is because of the acceleration, which is produced due to the force of gravity. In this topic, we will discuss acceleration due to Gravity formula. Let us learn about acceleration due to
Acceleration due to gravity is the acceleration that is gained by an object due to the gravitational force. Its SI unit is ms². It has a magnitude as well as direction. Thus it is a vector quantity.
We represent acceleration due to gravity by the symbol g. Its standard value on the surface of the earth at sea level is 9.8 ms². Its computation formula is based on Newton’s Second Law of Motion and Newton’s Law of Universal Gravitation.
Acceleration Due to Gravity Formula
Near the surface of Earth, the acceleration due to gravity is approximately constant. But, at large distances from the Earth, or around other planets or moons, it is varying. The acceleration due to gravity depends on the terms as the following:
Mass of the body,
Distance from the center of mass,
Constant G i.e. Universal gravitational constant.
g =
G
M
r
2
Where,
g Acceleration due to gravity
(
u
n
i
t
s
m
s
−
1
)
G The universal gravitational constant,
=
6.673
×
10
−
11
N
m
2
K
g
2
m Mass of a very large body like Earth.
r The distance from the center of mass of the large body
Variation of g with Height:
Acceleration due to gravity varies with the height from the surface of the earth. Its computation can be done as follows:
g
h
=
g
(
1
+
h
R
)
−
2
Where,
g Acceleration due to gravity at the surface.
g
h
Acceleration due to gravity at the height h.
R The radius of the earth.
h Height from the earth’s surface.
It is clear that the value of g decreases with an increase in height of an object. Hence the value of g becomes zero at infinite distance from the earth.
S
Example-1: The radius of the moon is
1.74
×
10
6
m
. The mass of the moon is taken as
7.35
×
10
22
kg. Find out the acceleration due to gravity on the surface of the moon.
Solution: On the surface of the moon, the distance to the center of mass will be the same as the radius.
Thus, r =
1.74
×
10
6
m
.
Mass of the object i.e. moon,
m =
7.35
×
10
22
kg
As we know that, universal gravitational constant G =
6.673
×
10
−
11
The acceleration due to gravity on the surface of the moon can be computed by using the formula as below:
g =
G
M
r
2
Substituting the values,
g =
6.673
×
10
−
11
×
7.35
×
10
22
(
1.74
×
10
6
)
2
g =
1.620
m
s
−
2
Hence, value of the acceleration due to gravity is
1.620
m
s
−
2