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A planet in a distant solar system is 10 times more massive than the earth and its radius is 10 times smaller. Given that the escape velocity from the earth is 11 km s⁻¹, the escape velocity from the surface of the planet would be
The escape velocity vₑ of a body depends on the gravitational pull of the celestial body it is escaping from. For Earth, the escape velocity is derived based on its mass M and radius R using the formula: vₑ ∝ √((2 GM)/R) On a different planet, if its mass is 10 times that of Earth and its radius isRead more
The escape velocity vₑ of a body depends on the gravitational pull of the celestial body it is escaping from. For Earth, the escape velocity is derived based on its mass M and radius R using the formula:
vₑ ∝ √((2 GM)/R)
On a different planet, if its mass is 10 times that of Earth and its radius is 1/10th of Earth’s, the escape velocity will increase significantly. Substituting the planet’s properties into the formula reveals that the escape velocity is proportional to:
vₑ ∝ 10 x √((2 GM)/R)
This results in an escape velocity that is 10 times that of Earth, making it approximately 110 km/s.
vₑ(earth) = √((2 GM)/R) = 11 km s⁻¹
See lessvₑ(planet) = √((2 G x 10 M)/R/10) = 10√((2 GM)/R)
= 10 x 11 = 110 km s⁻¹
The electric mains in the house are marked 220 V, 50Hz. Write down the equuation for instantaneous voltage.
The electric mains in a house marked 220 V and 50 Hz have an instantaneous voltage equation given by V =220√2 sin (2π(50)t) = 311sin(314t). Here, 311 V is the peak voltage, and 50 Hz is the frequency. For more visit here: https://www.tiwariacademy.com/ncert-solutions/class-12/physics/chapter-7/
The electric mains in a house marked 220 V and 50 Hz have an instantaneous voltage equation given by V =220√2 sin (2π(50)t) = 311sin(314t). Here, 311 V is the peak voltage, and 50 Hz is the frequency.
For more visit here:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-12/physics/chapter-7/
Two spherical bodies of mass M and 5 M and radii R and 2R respectively are released in free space with initial separation between their centers equal to 12 R. If they affact each other due to gravitational force only, then the distance covered by the smaller body just before collision is
The gravitational force between two spheres depends on their masses and the distance between their centers. The smaller sphere experiences a force due to the gravitational attraction of the larger sphere, which results in acceleration. The acceleration of the smaller sphere is calculated by dividingRead more
The gravitational force between two spheres depends on their masses and the distance between their centers. The smaller sphere experiences a force due to the gravitational attraction of the larger sphere, which results in acceleration. The acceleration of the smaller sphere is calculated by dividing the gravitational force by its mass.
Similarly, the larger sphere also experiences acceleration due to the gravitational force exerted by the smaller sphere. However, since the larger sphere has a greater mass, its acceleration is proportionally smaller. The relationship between the force, mass, and acceleration illustrates the mutual interaction governed by Newton’s law of gravitation.
Gravitational force between the two spheres,
See lessF = (GM x 5 M)/(12 R – x)²
Acceleration of smaller body,
a₁ = F/M = (G x 5 M)/(12 R – x)²
Acceleration of larger body,
a₂ = F/5 M = GM/(12 R – x)²
Suppose that the gravitational force varies inversely as the nth power of distance. Then, the times period of a planet in circular orbit of radius R around the sun will be proportional to
If the gravitational force varies inversely with the n-th power of the distance, the relationship between the gravitational force and the orbital radius changes accordingly. For a planet in a circular orbit, the centripetal force required for circular motion is provided by this gravitational force.Read more
If the gravitational force varies inversely with the n-th power of the distance, the relationship between the gravitational force and the orbital radius changes accordingly. For a planet in a circular orbit, the centripetal force required for circular motion is provided by this gravitational force. The balance of these forces determines the planet’s orbital velocity.
The time period of the orbit depends on the radius of the orbit and the orbital velocity. By analyzing this relationship under the modified gravitational law, it can be shown that the time period of the planet’s orbit is proportional to R⁽ⁿ ⁺ ¹⁾/². This reflects the dependence of orbital dynamics on the nature of the gravitational force.
See lessThe change in the value of g at a height at a depth d below the surface of earth. When both d and h are much smaller than the radius of earth, then which one of the following is correct?
When analyzing the change in gravitational acceleration g at a height h above or a depth d below the Earth's surface, the variations depend on their relationship to the Earth's radius R. For both h and d much smaller than R, the following approximations hold: 1. At a height h above the surface, g deRead more
When analyzing the change in gravitational acceleration g at a height h above or a depth d below the Earth’s surface, the variations depend on their relationship to the Earth’s radius R. For both h and d much smaller than R, the following approximations hold:
1. At a height h above the surface, g decreases proportionally to 1 – 2h/R, due to the inverse square law of gravitation.
2. At a depth d below the surface, g decreases proportionally to 1 – d/R, because the effective mass contributing to gravity reduces linearly with depth.
If d = 2h, the proportional reduction in g at height h and depth d would be equivalent, demonstrating a symmetry in the changes under these conditions.
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