While ejecting a body from the Earth, escape velocity becomes very crucial to overcome gravitational force so that it becomes free from the planet's influence. The conventional value of escape velocity from Earth's surface is about 11.2 km/s. But if the launch angle is considered to be 60 degrees wiRead more
While ejecting a body from the Earth, escape velocity becomes very crucial to overcome gravitational force so that it becomes free from the planet’s influence. The conventional value of escape velocity from Earth’s surface is about 11.2 km/s. But if the launch angle is considered to be 60 degrees with the vertical, then the whole speed needed for escape becomes a different story.
The required launch velocity, at an angle, would be the summation of the components of velocity. The vertical component needs to equal the escape velocity so that the body can rise infinitely against the gravity of Earth. Now, for an angle of 60 degrees, we find that in relation between the required launch velocity and the escape velocity, it shows that the necessary velocity for escape is increasing.
Specifically, when launched at an angle of 60 degrees, the total velocity needed becomes approximately 11√3 km/s. This is because the vertical component of the launch velocity has to compensate for the angle at which the body is launched. Consequently, a launch speed of 11√3 km/s guarantees that the vertical component will match the required escape velocity of 11.2 km/s and the body will break free from Earth’s gravitational influence. This is how launch angle affects the dynamics of reaching escape velocity.
Escape velocity is defined as the minimum speed for an object to break away from the gravitational pull of a celestial body, say Earth. For Earth, this escape velocity is around 11.2 km/s. If a body is projected vertically upwards, reaching this speed enables it to overcome gravitational attractionRead more
Escape velocity is defined as the minimum speed for an object to break away from the gravitational pull of a celestial body, say Earth. For Earth, this escape velocity is around 11.2 km/s. If a body is projected vertically upwards, reaching this speed enables it to overcome gravitational attraction without needing any further propulsion.
In the case of launching at an angle, the scenario is different. If the projection is made at an angle of 45 degrees with respect to the vertical, then the launch velocity would need to be adjusted so that the vertical component was equal to the escape velocity at the given point. With the Earth’s gravity present, this would ensure indefinite ascent of the body projected. For an angle of 45 degrees, a relation between the total velocity at launch and the escape velocity is critical.
More precisely, the launching speed required now becomes 11.2√2 km/s when launched at this angle of projection. The adjustment in this case will take into account both the need to maintain enough momentum upwards and the angle of projection. By attaining this higher speed, it ensures that the vertical component of the velocity is high enough to escape Earth’s gravitational pull. This shows how the launching angles impact significantly on the dynamics involved in achieving escape velocity.
Escape velocity is the minimum speed an object must reach to break free from a celestial body's gravitational influence. For Earth, the escape velocity is approximately 11.2 km/s. This value is determined by the mass and radius of the planet. When considering changes to these parameters, the escapeRead more
Escape velocity is the minimum speed an object must reach to break free from a celestial body’s gravitational influence. For Earth, the escape velocity is approximately 11.2 km/s. This value is determined by the mass and radius of the planet. When considering changes to these parameters, the escape velocity will also change.
If the mass of the Earth doubles while the radius is halved, the implications for escape velocity are significant. A doubling of the Earth’s mass increases the gravitational force exerted on an object at its surface. Concurrently, reducing the radius decreases the distance from the center of the Earth to the surface, which also increases the gravitational force felt by an object.
As a result of these changes, the new escape velocity becomes approximately 22.4 km/s. This is double the original escape velocity. Therefore, a body must now reach this higher speed to escape Earth’s gravitational pull. This scenario highlights the relationship between mass, radius, and escape velocity, illustrating how modifications in these parameters can lead to significant changes in the energy required for an object to break free from a planet’s gravitational field.
The escape velocity of a celestial body is the minimum speed required for an object to break free from its gravitational influence without any further propulsion. For a planet with a mass equal to that of Earth but with a radius that is one-fourth of Earth's radius, the escape velocity is much higheRead more
The escape velocity of a celestial body is the minimum speed required for an object to break free from its gravitational influence without any further propulsion. For a planet with a mass equal to that of Earth but with a radius that is one-fourth of Earth’s radius, the escape velocity is much higher than that of Earth.
This is due to the relation between mass and radius in gravitational physics. The gravitational pull an object experiences on the surface of the planet depends on both the mass of the planet and the distance from its center. As the radius is reduced to one-fourth, the gravitational force at the surface increases, and it needs a greater speed to escape the gravitational field of the planet.
This means that the escape velocity for this hypothetical planet is about 22.4 km/s, which is double the escape velocity of Earth, which is about 11.2 km/s. This means that an object launched from the surface of this planet must achieve a much higher speed to overcome the stronger gravitational attraction. Understanding this concept is crucial for space missions and exploring the dynamics of celestial bodies in our universe.
When a planet revolves in an elliptical orbit around the sun, a variety of energies and angular momentums are involved. The total energy, E in this context, is thus very important since it stands for the sum of kinetic energy T and gravitational potential energy V of the planet. In any bound elliptiRead more
When a planet revolves in an elliptical orbit around the sun, a variety of energies and angular momentums are involved. The total energy, E in this context, is thus very important since it stands for the sum of kinetic energy T and gravitational potential energy V of the planet.
In any bound elliptical orbit of a planet, total energy is always negative. This negative value means that the planet is gravitationally bound to the sun, with insufficient energy to leave the influence of the gravitational pull from the sun. The reason why the gravitational potential energy is negative is because of the attraction caused by gravity; the more negative this energy gets as the planet approaches the sun, the greater is its kinetic energy, which peaks at periapsis, the closest point of the orbit to the sun.
On the other hand, at apoapsis, where the planet is farthest from the sun, its kinetic energy is lower, but the gravitational potential energy is still negative. The total energy is consistently negative, and this is what reflects the stability of the orbit. Therefore, it is evident that for a planet in an elliptical orbit, the total energy is always negative, which is a further proof of the concept of gravitational binding in celestial mechanics.
The period of revolution of a planet from the Sun is related to its distance from the Sun by Kepler's third law. This law explains that the square of the orbital period of a planet is proportional to the cube of its average distance from the Sun. Here, planet A takes 8 times longer to complete one rRead more
The period of revolution of a planet from the Sun is related to its distance from the Sun by Kepler’s third law. This law explains that the square of the orbital period of a planet is proportional to the cube of its average distance from the Sun. Here, planet A takes 8 times longer to complete one revolution around the Sun as compared to planet B.
Using Kepler’s law, the relationship between the orbital period and distance implies that if the period of revolution of planet A is 8 times that of planet B, the cube root of the square of this ratio will give the ratio of their distances from the Sun. Simplifying this relationship, it is determined that the distance of planet A from the Sun is 4 times greater than that of planet B.
This result demonstrates how much more time is taken by the planet to complete its orbit as the distance from the Sun increases. The universality of this principle among all celestial bodies that move in elliptical orbits tells us a lot about planetary systems’ structure and dynamics, bringing to light harmony between orbital period and distance in celestial mechanics.
Given the mass of the sphere M = 100 kg, the mass of the particle m = 10 g or 10⁻² kg, and the radius R = 10 cm or 0.01 m, the initial potential energy (P.E.) of the two bodies is calculated based on their gravitational interaction. When the particle is moved far away from the sphere, the gravitatioRead more
Given the mass of the sphere M = 100 kg, the mass of the particle m = 10 g or 10⁻² kg, and the radius R = 10 cm or 0.01 m, the initial potential energy (P.E.) of the two bodies is calculated based on their gravitational interaction. When the particle is moved far away from the sphere, the gravitational potential energy of the system becomes zero. The work done in moving the particle from its initial position to an infinite distance is the difference between the final and initial potential energies, resulting in a positive value.
In the given scenario, the initial potential energy of the system is determined by the gravitational interaction between the two masses. When the particle is moved infinitely far away, the potential energy becomes zero. To move the particle from its initial position to infinity, the kinetic energy rRead more
In the given scenario, the initial potential energy of the system is determined by the gravitational interaction between the two masses. When the particle is moved infinitely far away, the potential energy becomes zero. To move the particle from its initial position to infinity, the kinetic energy required is equivalent to the change in potential energy. This kinetic energy can be understood as the work needed to overcome the gravitational attraction between the two bodies. The amount of kinetic energy required is proportional to the gravitational acceleration, the mass of the particle, and the radius of the sphere.
The gravitational potential energy (P.E.) of a mass m in an orbit of radius R is negative and proportional to the inverse of the radius. When the radius is doubled, the potential energy becomes less negative, reflecting the decrease in gravitational attraction. Similarly, when the radius is tripled,Read more
The gravitational potential energy (P.E.) of a mass m in an orbit of radius R is negative and proportional to the inverse of the radius. When the radius is doubled, the potential energy becomes less negative, reflecting the decrease in gravitational attraction. Similarly, when the radius is tripled, the potential energy further decreases. The change in gravitational potential energy between two orbits, such as from a radius of 2R to 3R, can be found by calculating the difference between the two potential energies. This difference represents the work done by the gravitational force in moving the object between these orbits.
If g is the acceleration due to gravity on the Earth's surface, the gain in potential energy of an object of mass m, when raised from the surface of the Earth to a height equal to the Earth's radius R, can be calculated by considering the change in gravitational potential energy. The potential energRead more
If g is the acceleration due to gravity on the Earth’s surface, the gain in potential energy of an object of mass m, when raised from the surface of the Earth to a height equal to the Earth’s radius R, can be calculated by considering the change in gravitational potential energy. The potential energy at the Earth’s surface and at the height R are compared, and the difference gives the gain in potential energy. This results in a gain of half the product of the object’s mass, gravitational acceleration, and the radius of the Earth.
For a satellite velocity from earth is 11.2 km/s. If a body is to be launched at an angle of 60ᵒ with the vertical, then escape velocity will be
While ejecting a body from the Earth, escape velocity becomes very crucial to overcome gravitational force so that it becomes free from the planet's influence. The conventional value of escape velocity from Earth's surface is about 11.2 km/s. But if the launch angle is considered to be 60 degrees wiRead more
While ejecting a body from the Earth, escape velocity becomes very crucial to overcome gravitational force so that it becomes free from the planet’s influence. The conventional value of escape velocity from Earth’s surface is about 11.2 km/s. But if the launch angle is considered to be 60 degrees with the vertical, then the whole speed needed for escape becomes a different story.
The required launch velocity, at an angle, would be the summation of the components of velocity. The vertical component needs to equal the escape velocity so that the body can rise infinitely against the gravity of Earth. Now, for an angle of 60 degrees, we find that in relation between the required launch velocity and the escape velocity, it shows that the necessary velocity for escape is increasing.
Specifically, when launched at an angle of 60 degrees, the total velocity needed becomes approximately 11√3 km/s. This is because the vertical component of the launch velocity has to compensate for the angle at which the body is launched. Consequently, a launch speed of 11√3 km/s guarantees that the vertical component will match the required escape velocity of 11.2 km/s and the body will break free from Earth’s gravitational influence. This is how launch angle affects the dynamics of reaching escape velocity.
See lessThe escape velocity from earth is 11.2 km/s. If a body is to be projected in a direction making an angle 45° to the vertical, then the escape velocity is
Escape velocity is defined as the minimum speed for an object to break away from the gravitational pull of a celestial body, say Earth. For Earth, this escape velocity is around 11.2 km/s. If a body is projected vertically upwards, reaching this speed enables it to overcome gravitational attractionRead more
Escape velocity is defined as the minimum speed for an object to break away from the gravitational pull of a celestial body, say Earth. For Earth, this escape velocity is around 11.2 km/s. If a body is projected vertically upwards, reaching this speed enables it to overcome gravitational attraction without needing any further propulsion.
In the case of launching at an angle, the scenario is different. If the projection is made at an angle of 45 degrees with respect to the vertical, then the launch velocity would need to be adjusted so that the vertical component was equal to the escape velocity at the given point. With the Earth’s gravity present, this would ensure indefinite ascent of the body projected. For an angle of 45 degrees, a relation between the total velocity at launch and the escape velocity is critical.
More precisely, the launching speed required now becomes 11.2√2 km/s when launched at this angle of projection. The adjustment in this case will take into account both the need to maintain enough momentum upwards and the angle of projection. By attaining this higher speed, it ensures that the vertical component of the velocity is high enough to escape Earth’s gravitational pull. This shows how the launching angles impact significantly on the dynamics involved in achieving escape velocity.
See lessThe escape velocity of a body on the surface of the earth is 11.2 km/s. If the earth’s mass increase to twice its present value and radius of the earth becomes half, the escape velocity becomes
Escape velocity is the minimum speed an object must reach to break free from a celestial body's gravitational influence. For Earth, the escape velocity is approximately 11.2 km/s. This value is determined by the mass and radius of the planet. When considering changes to these parameters, the escapeRead more
Escape velocity is the minimum speed an object must reach to break free from a celestial body’s gravitational influence. For Earth, the escape velocity is approximately 11.2 km/s. This value is determined by the mass and radius of the planet. When considering changes to these parameters, the escape velocity will also change.
If the mass of the Earth doubles while the radius is halved, the implications for escape velocity are significant. A doubling of the Earth’s mass increases the gravitational force exerted on an object at its surface. Concurrently, reducing the radius decreases the distance from the center of the Earth to the surface, which also increases the gravitational force felt by an object.
As a result of these changes, the new escape velocity becomes approximately 22.4 km/s. This is double the original escape velocity. Therefore, a body must now reach this higher speed to escape Earth’s gravitational pull. This scenario highlights the relationship between mass, radius, and escape velocity, illustrating how modifications in these parameters can lead to significant changes in the energy required for an object to break free from a planet’s gravitational field.
See lessFor a planet having mass equal to mass of the earth, the radius is one fourth of radius of the earth. Then escape velocity for this planet will be
The escape velocity of a celestial body is the minimum speed required for an object to break free from its gravitational influence without any further propulsion. For a planet with a mass equal to that of Earth but with a radius that is one-fourth of Earth's radius, the escape velocity is much higheRead more
The escape velocity of a celestial body is the minimum speed required for an object to break free from its gravitational influence without any further propulsion. For a planet with a mass equal to that of Earth but with a radius that is one-fourth of Earth’s radius, the escape velocity is much higher than that of Earth.
This is due to the relation between mass and radius in gravitational physics. The gravitational pull an object experiences on the surface of the planet depends on both the mass of the planet and the distance from its center. As the radius is reduced to one-fourth, the gravitational force at the surface increases, and it needs a greater speed to escape the gravitational field of the planet.
This means that the escape velocity for this hypothetical planet is about 22.4 km/s, which is double the escape velocity of Earth, which is about 11.2 km/s. This means that an object launched from the surface of this planet must achieve a much higher speed to overcome the stronger gravitational attraction. Understanding this concept is crucial for space missions and exploring the dynamics of celestial bodies in our universe.
See lessA planet is moving in an elliptical orbit around the sun. If T.V and L stands respectively for its kinetic energy, gravitational potential energy, total energy and magnitude of angular momentum about the centre of force, which of the following is correct?
When a planet revolves in an elliptical orbit around the sun, a variety of energies and angular momentums are involved. The total energy, E in this context, is thus very important since it stands for the sum of kinetic energy T and gravitational potential energy V of the planet. In any bound elliptiRead more
When a planet revolves in an elliptical orbit around the sun, a variety of energies and angular momentums are involved. The total energy, E in this context, is thus very important since it stands for the sum of kinetic energy T and gravitational potential energy V of the planet.
In any bound elliptical orbit of a planet, total energy is always negative. This negative value means that the planet is gravitationally bound to the sun, with insufficient energy to leave the influence of the gravitational pull from the sun. The reason why the gravitational potential energy is negative is because of the attraction caused by gravity; the more negative this energy gets as the planet approaches the sun, the greater is its kinetic energy, which peaks at periapsis, the closest point of the orbit to the sun.
On the other hand, at apoapsis, where the planet is farthest from the sun, its kinetic energy is lower, but the gravitational potential energy is still negative. The total energy is consistently negative, and this is what reflects the stability of the orbit. Therefore, it is evident that for a planet in an elliptical orbit, the total energy is always negative, which is a further proof of the concept of gravitational binding in celestial mechanics.
See lessThe period of revolution of planet A around the sun is 8 times that of B. The distance of A from the sun is how many times greater than that of B from the sun?
The period of revolution of a planet from the Sun is related to its distance from the Sun by Kepler's third law. This law explains that the square of the orbital period of a planet is proportional to the cube of its average distance from the Sun. Here, planet A takes 8 times longer to complete one rRead more
The period of revolution of a planet from the Sun is related to its distance from the Sun by Kepler’s third law. This law explains that the square of the orbital period of a planet is proportional to the cube of its average distance from the Sun. Here, planet A takes 8 times longer to complete one revolution around the Sun as compared to planet B.
Using Kepler’s law, the relationship between the orbital period and distance implies that if the period of revolution of planet A is 8 times that of planet B, the cube root of the square of this ratio will give the ratio of their distances from the Sun. Simplifying this relationship, it is determined that the distance of planet A from the Sun is 4 times greater than that of planet B.
This result demonstrates how much more time is taken by the planet to complete its orbit as the distance from the Sun increases. The universality of this principle among all celestial bodies that move in elliptical orbits tells us a lot about planetary systems’ structure and dynamics, bringing to light harmony between orbital period and distance in celestial mechanics.
See lessA particle of mass 10g is kept on the surface of a uniform sphere of mass 100 kg and radius 10 cm. Find the work to be done against the gravitational force between them to take the particle far away from the sphere. You may take G = 6.67 x 10⁻¹¹ Nm²kg ⁻²
Given the mass of the sphere M = 100 kg, the mass of the particle m = 10 g or 10⁻² kg, and the radius R = 10 cm or 0.01 m, the initial potential energy (P.E.) of the two bodies is calculated based on their gravitational interaction. When the particle is moved far away from the sphere, the gravitatioRead more
Given the mass of the sphere M = 100 kg, the mass of the particle m = 10 g or 10⁻² kg, and the radius R = 10 cm or 0.01 m, the initial potential energy (P.E.) of the two bodies is calculated based on their gravitational interaction. When the particle is moved far away from the sphere, the gravitational potential energy of the system becomes zero. The work done in moving the particle from its initial position to an infinite distance is the difference between the final and initial potential energies, resulting in a positive value.
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See lessThe kinetic energy needed to project a body of mass m from the earth’s surface (radius R) to infinity is
In the given scenario, the initial potential energy of the system is determined by the gravitational interaction between the two masses. When the particle is moved infinitely far away, the potential energy becomes zero. To move the particle from its initial position to infinity, the kinetic energy rRead more
In the given scenario, the initial potential energy of the system is determined by the gravitational interaction between the two masses. When the particle is moved infinitely far away, the potential energy becomes zero. To move the particle from its initial position to infinity, the kinetic energy required is equivalent to the change in potential energy. This kinetic energy can be understood as the work needed to overcome the gravitational attraction between the two bodies. The amount of kinetic energy required is proportional to the gravitational acceleration, the mass of the particle, and the radius of the sphere.
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See lessEnergy required to move a body of mass m from an orbit of radius 2 R to 3 R is
The gravitational potential energy (P.E.) of a mass m in an orbit of radius R is negative and proportional to the inverse of the radius. When the radius is doubled, the potential energy becomes less negative, reflecting the decrease in gravitational attraction. Similarly, when the radius is tripled,Read more
The gravitational potential energy (P.E.) of a mass m in an orbit of radius R is negative and proportional to the inverse of the radius. When the radius is doubled, the potential energy becomes less negative, reflecting the decrease in gravitational attraction. Similarly, when the radius is tripled, the potential energy further decreases. The change in gravitational potential energy between two orbits, such as from a radius of 2R to 3R, can be found by calculating the difference between the two potential energies. This difference represents the work done by the gravitational force in moving the object between these orbits.
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See lessIf g is the acceleration due to gravity on the earth’s surface, the gain in the potential energy of an object of mass m raised from the surface of the earth to a height equal to the radius R of the earth, is
If g is the acceleration due to gravity on the Earth's surface, the gain in potential energy of an object of mass m, when raised from the surface of the Earth to a height equal to the Earth's radius R, can be calculated by considering the change in gravitational potential energy. The potential energRead more
If g is the acceleration due to gravity on the Earth’s surface, the gain in potential energy of an object of mass m, when raised from the surface of the Earth to a height equal to the Earth’s radius R, can be calculated by considering the change in gravitational potential energy. The potential energy at the Earth’s surface and at the height R are compared, and the difference gives the gain in potential energy. This results in a gain of half the product of the object’s mass, gravitational acceleration, and the radius of the Earth.
Check out for more :- https://www.tiwariacademy.com/ncert-solutions/class-11/physics/chapter-7/
See less