We can then solve for how kinetic energy would have increased for that increase of momentum by 20%. So first we give out the momentum and kinetic energy formula. p = mv, whereas K.E. = (1/2) mv². Express the Kinetic Energy as a Function of Momentum: Let's convert Kinetic energy as a function of momeRead more
We can then solve for how kinetic energy would have increased for that increase of momentum by 20%. So first we give out the momentum and kinetic energy formula.
p = mv, whereas K.E. = (1/2) mv².
Express the Kinetic Energy as a Function of Momentum:
Let’s convert Kinetic energy as a function of momentum; since momentum was multiplied by constant so was velocity by same amount:
K.E. = (1/2) mv = (1/2)(p/v).
From p = mv, we have:
v = p/m
Substituting v in the kinetic energy formula:
K.E. = (1/2) m (p/m)²
K.E. = (1/2) (p²/m)
Step 3: Calculate the initial and final momentum
Let the initial momentum be p. If momentum increases by 20%, the new momentum (p’) is:
p’ = p + 0.2p = 1.2p
Step 4: Calculate initial and final kinetic energy
To calculate the kinetic energy of a body given its mass and momentum, we can make use of the relationship between momentum (p) and kinetic energy (K.E.). Step 1: Write the formulas - Momentum (p) is given by: p = mv - Kinetic energy (K.E.) is given by: K.E. = (1/2) mv² Step 2: Relate momentum to kiRead more
To calculate the kinetic energy of a body given its mass and momentum, we can make use of the relationship between momentum (p) and kinetic energy (K.E.).
Step 1: Write the formulas
– Momentum (p) is given by:
p = mv
– Kinetic energy (K.E.) is given by:
K.E. = (1/2) mv²
Step 2: Relate momentum to kinetic energy
We can write kinetic energy in terms of momentum:
Using the relation p = mv, we can rearrange to find v:
v = p/m
Substituting this into the kinetic energy formula:
K.E. = (1/2) m (p/m)²
K.E. = (1/2) (p²/m)
Step 3: Substitute the values
Given:
– Mass (m) = 2 kg
– Momentum (p) = 2 N·s (Note: momentum is in kg·m/s which is equivalent to N·s)
Now substituting into the formula for kinetic energy:
To find the motion of the body placed on a rough horizontal plane, we need to consider the effects of friction and applied force. Given: Mass of the body (m) = 2 kg Coefficient of friction (μ) = 0.2 Acceleration due to gravity (g) = 9.8 m/s² (we will use 10 m/s² for simplicity) Step 1: Calculate theRead more
To find the motion of the body placed on a rough horizontal plane, we need to consider the effects of friction and applied force.
Given:
Mass of the body (m) = 2 kg
Coefficient of friction (μ) = 0.2
Acceleration due to gravity (g) = 9.8 m/s² (we will use 10 m/s² for simplicity)
Step 1: Calculate the normal force (N)
On a horizontal plane, the normal force (N) is equal to the weight of the body:
N = m * g
N = 2 kg * 10 m/s²
N = 20 N
Step 2: Find the maximum static friction (f_s)
The maximum static friction is given by:
f_s = μ * N
f_s = 0.2 * 20 N
f_s = 4 N
Step 3: Analyze the applied force (F)
Case (a): If F = 5 N
– The applied force (5 N) is greater than the maximum static friction (4 N).
– Hence, the body will move in the forward direction.
Case (b): If F = 3 N
– The applied force (3 N) is less than the maximum static friction (4 N).
– Thus, the body will not move and will remain at rest.
**Case (c):** If F = 3 N
– Since the applied force is not enough to overcome friction, the body will remain at rest.
Conclusion:
– For F = 5 N, the body will move in the forward direction.
– For F = 3 N, the body will remain at rest.
– Hence, both (a) and (c) are correct.
Final Answer:
The correct options are both (a) and (c).
When a body is moving in a circular path with constant speed, we have to look into the consequences of circular motion. 1. Work done will be zero: - True. Since the force (centripetal force) is always perpendicular to the direction of motion, no work is done on the body. 2. Acceleration will be zeroRead more
When a body is moving in a circular path with constant speed, we have to look into the consequences of circular motion.
1. Work done will be zero:
– True. Since the force (centripetal force) is always perpendicular to the direction of motion, no work is done on the body.
2. Acceleration will be zero:
– False. The velocity has a constant speed but is changing direction all the time, and hence there is a centripetal acceleration directed toward the center of the circular path.
3. No force acts on the body:
– False. A net force (centripetal force) is necessary to keep the body moving in a circular path and is directed towards the center.
4. Its velocity remains constant:
– False. True, the velocity magnitude is unchanged, but the velocity is a vector: its direction must change; it is not a constant.
To find the ratio of the potential energies of two springs with different spring constants when stretched by the same force, we can use the formula for elastic potential energy: U = (1/2) k x² Where: - U is the potential energy - k is the spring constant - x is the extension of the spring Step 1: ReRead more
To find the ratio of the potential energies of two springs with different spring constants when stretched by the same force, we can use the formula for elastic potential energy:
U = (1/2) k x²
Where:
– U is the potential energy
– k is the spring constant
– x is the extension of the spring
Step 1: Relationship between force and extension
According to Hooke’s Law, the force applied to a spring is related to the spring constant and the extension by:
F = kx
Rearranging gives us:
x = F/k
Step 2: Calculate potential energy for both springs
For spring 1 (k₁ = 1500 N/m):
U₁ = (1/2) k₁ x₁²
Substituting for x₁:
U₁ = (1/2) k₁ (F/k₁)²
U₁ = (1/2) k₁ (F²/k₁²)
U₁ = (1/2) (F²/k₁)
For spring 2 (k₂ = 3000 N/m):
U₂ = (1/2) k₂ x₂²
Substituting for x₂:
U₂ = (1/2) k₂ (F/k₂)²
U₂ = (1/2) k₂ (F²/k₂²)
U₂ = (1/2) (F²/k₂)
To find the ratio of the linear momentum of two particles with the same kinetic energy, we can use the formulas for kinetic energy and momentum. Step 1: Write the formula for kinetic energy The kinetic energy (K.E.) is given by: K.E. = (1/2) m v² Where: - m = mass - v = velocity Step 2: Set the kineRead more
To find the ratio of the linear momentum of two particles with the same kinetic energy, we can use the formulas for kinetic energy and momentum.
The acceleration due to gravity varies from the poles to the equator, mainly due to two factors: Earth's rotation and its shape. Earth is not a perfect sphere; it is an oblate spheroid, meaning it is slightly flattened at the poles and bulging at the equator. Consequently, the distance from the centRead more
The acceleration due to gravity varies from the poles to the equator, mainly due to two factors: Earth’s rotation and its shape. Earth is not a perfect sphere; it is an oblate spheroid, meaning it is slightly flattened at the poles and bulging at the equator. Consequently, the distance from the center of the Earth to the surface is greater at the equator than at the poles.
The force of gravity is strongest at the poles where the Earth’s rotation is least effective. This is because the gravitational force varies directly with the mass of the Earth and inversely with the square of the distance from the center of the Earth. Conversely, at the equator, the centrifugal force created by the Earth’s rotation opposes the force of gravity, thus reducing the effective acceleration due to gravity. This centrifugal force is greatest at the equator because the linear speed due to rotation is the highest at the equator.
Because of this, the acceleration due to gravity decreases from about 9.83 m/s² at the poles to about 9.78 m/s² at the equator. This variation is important in many applications such as navigation, satellite positioning, and understanding the behavior of objects in different gravitational fields.
The gravitational interaction between two bodies, such as the Earth and the Moon, is described by Newton's law of universal gravitation. According to this principle, every mass exerts a gravitational force on every other mass, and this force is proportional to the product of their masses and inverseRead more
The gravitational interaction between two bodies, such as the Earth and the Moon, is described by Newton’s law of universal gravitation. According to this principle, every mass exerts a gravitational force on every other mass, and this force is proportional to the product of their masses and inversely proportional to the square of the distance between their centers.
In the case of the Earth and the Moon, even though the Moon has only about 1% of the Earth’s mass, the gravitational pull that the Earth exerts on the Moon and the gravitational pull that the Moon exerts on the Earth are equal in magnitude. This equality is a direct consequence of Newton’s third law of motion, which states that for every action, there is an equal and opposite reaction. Therefore, the gravitational force that the Earth exerts on the Moon is matched by the gravitational force that the Moon exerts on the Earth, leading to a ratio of 1:1.
This relationship holds true regardless of the difference in masses. While the forces are equal, their effects are different due to the significant difference in mass. The Earth, being much more massive, has a stronger gravitational influence, resulting in the Moon’s orbit around it. However, the underlying principle remains that the gravitational forces between the two bodies are equal.
To understand the relationship between the speeds of two satellites in circular orbits around the Earth, we need to consider their orbital radii. In the given problem, satellite A orbits at radius 4R and travels at a speed of 3V, whereas satellite B orbits at radius R. The speed of a satellite in aRead more
To understand the relationship between the speeds of two satellites in circular orbits around the Earth, we need to consider their orbital radii. In the given problem, satellite A orbits at radius 4R and travels at a speed of 3V, whereas satellite B orbits at radius R.
The speed of a satellite in a circular orbit is dependent on the gravitational pull of the Earth and the radius of its orbit. In general, satellites closer to the Earth have a stronger gravitational force and therefore move at higher speeds. On the other hand, satellites that are farther away have slower speeds.
In this case, we know that satellite B has a radius of R and is much nearer the Earth than satellite A that orbits at 4R. We therefore see that satellite A was moving at 3V, so this immediately tells us how to compute the speed of the motion of satellite B since, having a smaller orbit radius, it will also be orbiting faster. Thus, the speed of satellite B is calculated to be 6V, which is twice the speed of satellite A. This shows the relationship between orbital radius and speed, which indicates that closer satellites must travel faster to maintain their orbits.
If the Earth were to lose its gravity, then a body near the Earth's surface would face drastic and immediate consequences. Gravity is the force that pulls objects toward the center of the Earth, keeping them grounded. Without this force, a body would no longer experience weight and would enter a staRead more
If the Earth were to lose its gravity, then a body near the Earth’s surface would face drastic and immediate consequences. Gravity is the force that pulls objects toward the center of the Earth, keeping them grounded. Without this force, a body would no longer experience weight and would enter a state of free fall.
In this scenario, everything ranging from individuals, houses, and other moving vehicles would start levitating in space and going off the Earth’s crust. This would be caused by a complete absence of gravity force. There is no pull to keep anchored any objects on the earth surface. Anything in motion will continue flowing without being halted by forces like gravity because it never acted in a pulling direction during the previous instance.
Moreover, it would have devastating consequences on the Earth’s atmosphere because gravity holds it there, and without it, air would dissipate into space, and people as well as other living things cannot breathe. Loss of gravity will affect the structural integrity of buildings and other structures leading to possible catastrophic failure.
In a nutshell, if Earth lost its gravity, all bodies would float into space, and life on this earth would be altered so much that it would never be the same, rendering the environment uninhabitable.
If momentum is increased by 20 %, then kinetic energy increases by
We can then solve for how kinetic energy would have increased for that increase of momentum by 20%. So first we give out the momentum and kinetic energy formula. p = mv, whereas K.E. = (1/2) mv². Express the Kinetic Energy as a Function of Momentum: Let's convert Kinetic energy as a function of momeRead more
We can then solve for how kinetic energy would have increased for that increase of momentum by 20%. So first we give out the momentum and kinetic energy formula.
p = mv, whereas K.E. = (1/2) mv².
Express the Kinetic Energy as a Function of Momentum:
Let’s convert Kinetic energy as a function of momentum; since momentum was multiplied by constant so was velocity by same amount:
K.E. = (1/2) mv = (1/2)(p/v).
From p = mv, we have:
v = p/m
Substituting v in the kinetic energy formula:
K.E. = (1/2) m (p/m)²
K.E. = (1/2) (p²/m)
Step 3: Calculate the initial and final momentum
Let the initial momentum be p. If momentum increases by 20%, the new momentum (p’) is:
p’ = p + 0.2p = 1.2p
Step 4: Calculate initial and final kinetic energy
Initial K.E. (K.E.₁):
K.E.₁ = (1/2) (p²/m)
Final K.E. (K.E.₂):
K.E.₂ = (1/2) (p’²/m)
K.E.₂ = (1/2) [(1.2p)²/m]
K.E.₂ = (1/2) [1.44p²/m]
Step 5: Find the increase in kinetic energy
Now, we calculate the increase in kinetic energy:
Increase in K.E. = K.E.₂ – K.E.₁
Increase in K.E. = (1/2) [1.44p²/m] – (1/2) [p²/m]
Increase in K.E. = (1/2m) [1.44p² – p²]
Increase in K.E. = (1/2m) [0.44p²]
Percentage increase in K.E. is given by:
Percentage increase = (Increase in K.E. / K.E.₁) × 100
Percentage increase = [(0.44p²/2m) / (p²/2m)] × 100
Percentage increase = 0.44 × 100
Percentage increase = 44%
Final Answer:
See lessIf the momentum is increased by 20%, then it increases the Kinetic energy with 44%.
The kinetic energy of body of mass 2 kg and momentum of 2 N is
To calculate the kinetic energy of a body given its mass and momentum, we can make use of the relationship between momentum (p) and kinetic energy (K.E.). Step 1: Write the formulas - Momentum (p) is given by: p = mv - Kinetic energy (K.E.) is given by: K.E. = (1/2) mv² Step 2: Relate momentum to kiRead more
To calculate the kinetic energy of a body given its mass and momentum, we can make use of the relationship between momentum (p) and kinetic energy (K.E.).
Step 1: Write the formulas
– Momentum (p) is given by:
p = mv
– Kinetic energy (K.E.) is given by:
K.E. = (1/2) mv²
Step 2: Relate momentum to kinetic energy
We can write kinetic energy in terms of momentum:
Using the relation p = mv, we can rearrange to find v:
v = p/m
Substituting this into the kinetic energy formula:
K.E. = (1/2) m (p/m)²
K.E. = (1/2) (p²/m)
Step 3: Substitute the values
Given:
– Mass (m) = 2 kg
– Momentum (p) = 2 N·s (Note: momentum is in kg·m/s which is equivalent to N·s)
Now substituting into the formula for kinetic energy:
K.E. = (1/2) (p²/m)
K.E. = (1/2) [(2)² / 2]
K.E. = (1/2) [4 / 2]
K.E. = (1/2) [2]
K.E. = 1 J
Final Answer:
The kinetic energy of the body is 1 J.
Click here:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-11/physics/chapter-5/
A body of mass 2 kg is placed on rough horizontal plane. The coefficient of friction between body and plane is 0.2 Then
To find the motion of the body placed on a rough horizontal plane, we need to consider the effects of friction and applied force. Given: Mass of the body (m) = 2 kg Coefficient of friction (μ) = 0.2 Acceleration due to gravity (g) = 9.8 m/s² (we will use 10 m/s² for simplicity) Step 1: Calculate theRead more
To find the motion of the body placed on a rough horizontal plane, we need to consider the effects of friction and applied force.
Given:
Mass of the body (m) = 2 kg
Coefficient of friction (μ) = 0.2
Acceleration due to gravity (g) = 9.8 m/s² (we will use 10 m/s² for simplicity)
Step 1: Calculate the normal force (N)
On a horizontal plane, the normal force (N) is equal to the weight of the body:
N = m * g
N = 2 kg * 10 m/s²
N = 20 N
Step 2: Find the maximum static friction (f_s)
The maximum static friction is given by:
f_s = μ * N
f_s = 0.2 * 20 N
f_s = 4 N
Step 3: Analyze the applied force (F)
Case (a): If F = 5 N
– The applied force (5 N) is greater than the maximum static friction (4 N).
– Hence, the body will move in the forward direction.
Case (b): If F = 3 N
– The applied force (3 N) is less than the maximum static friction (4 N).
– Thus, the body will not move and will remain at rest.
**Case (c):** If F = 3 N
– Since the applied force is not enough to overcome friction, the body will remain at rest.
Conclusion:
– For F = 5 N, the body will move in the forward direction.
– For F = 3 N, the body will remain at rest.
– Hence, both (a) and (c) are correct.
Final Answer:
The correct options are both (a) and (c).
Click here for more:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-11/physics/chapter-5/
When a body moves with constant speed in a circular path, then
When a body is moving in a circular path with constant speed, we have to look into the consequences of circular motion. 1. Work done will be zero: - True. Since the force (centripetal force) is always perpendicular to the direction of motion, no work is done on the body. 2. Acceleration will be zeroRead more
When a body is moving in a circular path with constant speed, we have to look into the consequences of circular motion.
1. Work done will be zero:
– True. Since the force (centripetal force) is always perpendicular to the direction of motion, no work is done on the body.
2. Acceleration will be zero:
– False. The velocity has a constant speed but is changing direction all the time, and hence there is a centripetal acceleration directed toward the center of the circular path.
3. No force acts on the body:
– False. A net force (centripetal force) is necessary to keep the body moving in a circular path and is directed towards the center.
4. Its velocity remains constant:
– False. True, the velocity magnitude is unchanged, but the velocity is a vector: its direction must change; it is not a constant.
Final Answer:
Work done will be zero.
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See lesshttps://www.tiwariacademy.com/ncert-solutions/class-11/physics/chapter-5/
Two springs of spring constant 1500 N/m and 3000 N/m respectively are stretched with a same force. Their potential energies will be in the ratio of
To find the ratio of the potential energies of two springs with different spring constants when stretched by the same force, we can use the formula for elastic potential energy: U = (1/2) k x² Where: - U is the potential energy - k is the spring constant - x is the extension of the spring Step 1: ReRead more
To find the ratio of the potential energies of two springs with different spring constants when stretched by the same force, we can use the formula for elastic potential energy:
U = (1/2) k x²
Where:
– U is the potential energy
– k is the spring constant
– x is the extension of the spring
Step 1: Relationship between force and extension
According to Hooke’s Law, the force applied to a spring is related to the spring constant and the extension by:
F = kx
Rearranging gives us:
x = F/k
Step 2: Calculate potential energy for both springs
For spring 1 (k₁ = 1500 N/m):
U₁ = (1/2) k₁ x₁²
Substituting for x₁:
U₁ = (1/2) k₁ (F/k₁)²
U₁ = (1/2) k₁ (F²/k₁²)
U₁ = (1/2) (F²/k₁)
For spring 2 (k₂ = 3000 N/m):
U₂ = (1/2) k₂ x₂²
Substituting for x₂:
U₂ = (1/2) k₂ (F/k₂)²
U₂ = (1/2) k₂ (F²/k₂²)
U₂ = (1/2) (F²/k₂)
Step 3: Find the ratio of potential energies
Now we can find the ratio of U₁ to U₂:
U₁ / U₂ = (F² / (2k₁)) / (F² / (2k₂))
U₁ / U₂ = k₂ / k₁
U₁ / U₂ = 3000 / 1500
U₁ / U₂ = 2
Final Answer:
The potential energies of the two springs will be in the ratio of 2 : 1.
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Kinetic energy of particles of mass 10 g and 40 g is same, the ratio of their linear momentum is
To find the ratio of the linear momentum of two particles with the same kinetic energy, we can use the formulas for kinetic energy and momentum. Step 1: Write the formula for kinetic energy The kinetic energy (K.E.) is given by: K.E. = (1/2) m v² Where: - m = mass - v = velocity Step 2: Set the kineRead more
To find the ratio of the linear momentum of two particles with the same kinetic energy, we can use the formulas for kinetic energy and momentum.
Step 1: Write the formula for kinetic energy
The kinetic energy (K.E.) is given by:
K.E. = (1/2) m v²
Where:
– m = mass
– v = velocity
Step 2: Set the kinetic energies equal
Let:
– m₁ = 10 g = 0.01 kg
– m₂ = 40 g = 0.04 kg
Since their kinetic energies are the same:
(1/2) m₁ v₁² = (1/2) m₂ v₂²
Cancelling (1/2) from both sides gives:
m₁ v₁² = m₂ v₂²
Step 3: Express velocity in terms of momentum
The momentum (p) is given by:
p = mv
From the above equation:
0.01 v₁² = 0.04 v₂²
Rearranging gives us:
v₁²/v₂² = 0.04/0.01
v₁²/v₂² = 4
Step 4: Find the ratio of velocities
Taking the square root of both sides:
v₁/v₂ = 2
Step 5: Calculate the ratio of momenta
Now, using the definition of momentum:
p₁ = m₁ v₁
p₂ = m₂ v₂
The ratio of their momenta is:
p₁/p₂ = (m₁ v₁) / (m₂ v₂)
p₁/p₂ = (0.01 v₁) / (0.04 v₂)
Substituting the ratio of velocities:
p₁/p₂ = (0.01 * 2v₂) / (0.04 v₂)
p₁/p₂ = (0.02 / 0.04)
p₁/p₂ = 1/2
Final Answer:
The ratio of their linear momentum is 1/2.
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Acceleration due to gravity
The acceleration due to gravity varies from the poles to the equator, mainly due to two factors: Earth's rotation and its shape. Earth is not a perfect sphere; it is an oblate spheroid, meaning it is slightly flattened at the poles and bulging at the equator. Consequently, the distance from the centRead more
The acceleration due to gravity varies from the poles to the equator, mainly due to two factors: Earth’s rotation and its shape. Earth is not a perfect sphere; it is an oblate spheroid, meaning it is slightly flattened at the poles and bulging at the equator. Consequently, the distance from the center of the Earth to the surface is greater at the equator than at the poles.
The force of gravity is strongest at the poles where the Earth’s rotation is least effective. This is because the gravitational force varies directly with the mass of the Earth and inversely with the square of the distance from the center of the Earth. Conversely, at the equator, the centrifugal force created by the Earth’s rotation opposes the force of gravity, thus reducing the effective acceleration due to gravity. This centrifugal force is greatest at the equator because the linear speed due to rotation is the highest at the equator.
Because of this, the acceleration due to gravity decreases from about 9.83 m/s² at the poles to about 9.78 m/s² at the equator. This variation is important in many applications such as navigation, satellite positioning, and understanding the behavior of objects in different gravitational fields.
See lessThe mass of moon is 1% mass of earth. The ratio of gravitational pull of earth on moon and that of moon on earth will be
The gravitational interaction between two bodies, such as the Earth and the Moon, is described by Newton's law of universal gravitation. According to this principle, every mass exerts a gravitational force on every other mass, and this force is proportional to the product of their masses and inverseRead more
The gravitational interaction between two bodies, such as the Earth and the Moon, is described by Newton’s law of universal gravitation. According to this principle, every mass exerts a gravitational force on every other mass, and this force is proportional to the product of their masses and inversely proportional to the square of the distance between their centers.
In the case of the Earth and the Moon, even though the Moon has only about 1% of the Earth’s mass, the gravitational pull that the Earth exerts on the Moon and the gravitational pull that the Moon exerts on the Earth are equal in magnitude. This equality is a direct consequence of Newton’s third law of motion, which states that for every action, there is an equal and opposite reaction. Therefore, the gravitational force that the Earth exerts on the Moon is matched by the gravitational force that the Moon exerts on the Earth, leading to a ratio of 1:1.
This relationship holds true regardless of the difference in masses. While the forces are equal, their effects are different due to the significant difference in mass. The Earth, being much more massive, has a stronger gravitational influence, resulting in the Moon’s orbit around it. However, the underlying principle remains that the gravitational forces between the two bodies are equal.
See lessThe radii of circular orbits of two satellites A and B of the earth are 4R and R, respectively. If the speed of satellite A is 3 V, the speed of satellite B will be
To understand the relationship between the speeds of two satellites in circular orbits around the Earth, we need to consider their orbital radii. In the given problem, satellite A orbits at radius 4R and travels at a speed of 3V, whereas satellite B orbits at radius R. The speed of a satellite in aRead more
To understand the relationship between the speeds of two satellites in circular orbits around the Earth, we need to consider their orbital radii. In the given problem, satellite A orbits at radius 4R and travels at a speed of 3V, whereas satellite B orbits at radius R.
The speed of a satellite in a circular orbit is dependent on the gravitational pull of the Earth and the radius of its orbit. In general, satellites closer to the Earth have a stronger gravitational force and therefore move at higher speeds. On the other hand, satellites that are farther away have slower speeds.
In this case, we know that satellite B has a radius of R and is much nearer the Earth than satellite A that orbits at 4R. We therefore see that satellite A was moving at 3V, so this immediately tells us how to compute the speed of the motion of satellite B since, having a smaller orbit radius, it will also be orbiting faster. Thus, the speed of satellite B is calculated to be 6V, which is twice the speed of satellite A. This shows the relationship between orbital radius and speed, which indicates that closer satellites must travel faster to maintain their orbits.
See lessIf the earth loses its gravity, then for a body
If the Earth were to lose its gravity, then a body near the Earth's surface would face drastic and immediate consequences. Gravity is the force that pulls objects toward the center of the Earth, keeping them grounded. Without this force, a body would no longer experience weight and would enter a staRead more
If the Earth were to lose its gravity, then a body near the Earth’s surface would face drastic and immediate consequences. Gravity is the force that pulls objects toward the center of the Earth, keeping them grounded. Without this force, a body would no longer experience weight and would enter a state of free fall.
In this scenario, everything ranging from individuals, houses, and other moving vehicles would start levitating in space and going off the Earth’s crust. This would be caused by a complete absence of gravity force. There is no pull to keep anchored any objects on the earth surface. Anything in motion will continue flowing without being halted by forces like gravity because it never acted in a pulling direction during the previous instance.
Moreover, it would have devastating consequences on the Earth’s atmosphere because gravity holds it there, and without it, air would dissipate into space, and people as well as other living things cannot breathe. Loss of gravity will affect the structural integrity of buildings and other structures leading to possible catastrophic failure.
In a nutshell, if Earth lost its gravity, all bodies would float into space, and life on this earth would be altered so much that it would never be the same, rendering the environment uninhabitable.
See less