Homework Question : Where will a body weigh more at equator or at pole? Explain why? HW Solution:
Earth is flattened at poles and bulged out at equator. Therefore, polar radius (Rp) is smaller than equatorial radius (Re). Well Done!!! In this we will be covering…...
Gravitational potential energy
Escape Velocity
Orbital velocity
Energy of satellite in Orbit Gravitational potential energy
Work done by system = - ΔU
The change in gravitational potential energy is
F(r) is the Gravitational Force. Gravitational potential energy
As always, the reference point where the potential energy is zero is arbitrary. Usually choose it at
This gives Gravitational potential energy
This is valid only for r ≥ RE & NOT for r < RE –That is, it is valid outside the Earth’s surface but NOT inside it!
–U is negative because of the choice of Ui At what height from the surface of earth the gravitation potential and the value of g are -5.4 × 107 J kg2 and 6.0 ms2 respectively? (take, the radius of earth as 6400 km)
a 2000 km
b 1600 km
c 1400 km
d 2600 km
Escape velocity
If an object of mass m projected upward from a point at a distance (h+RE ) from the centre of the earth. with an initial speed, Vi. Suppose the object did reach infinity and that its speed there was Vf . → The energy of an object is the sum of potential and kinetic energy.
Escape velocity
Using the relation , we get Escape velocity
The escape velocity of Earth is 11.19 km/s.
→ The escape velocity varies depending upon the object distance from centre of gravity. A satellite orbits the Earth with speed v in a circular orbit of radius r. At what radius would the satellite orbit with speed v/3?
a 3r
b 4r 2 V c 6r 4 V d 9r A satellite orbits the Earth with speed v in a circular orbit of radius r. At what radius would the satellite orbit with speed v/3?
a 3r
b 4r 2 V c 6r 4 V d 9r The escape velocity of a rocket on the earth is 11.2 km/sec. Its value on a planet where acceleration due to gravity is twice that on the earth and the diameter of the planet is twice that of the earth will be in (km/sec)
a 11.2 2 V b 5.6 4 V c 22.4
d 33.6 The escape velocity of a rocket on the earth is 11.2 km/sec. Its value on a planet where acceleration due to gravity is twice that on the earth and the diameter of the planet is twice that of the earth will be in (km/sec)
a 11.2 2 V b 5.6 4 V c 22.4
d 33.6 Escape velocity and Orbital velocity Satellite (Orbital velocity) Energy and Satellite
Consider a satellite of mass m moving with a speed v in the vicinity of a large mass M The total mechanical energy is the sum of the system’s kinetic and potential energies. E = K +U
•In a system in which m is bound in an orbit around M, can show that E must be less than 0 Energy of Satellite in orbit
The gravitational force supplies the centripetal force: 2 2 Fg = G(Mm/r ) = ma = m(v /r) Multiply both sides by r & divide by 2: G[(Mm)/(2r)] = (½)mv2 This gives, Energy of Satellite in orbit
→ The total mechanical energy is negative in for a circular orbit. → The kinetic energy is positive and is equal to half the absolute value of the potential energy The orbital velocity of an artificial satellite in a circular orbit just above the earth's surface is ν. For a satellite orbiting at an altitude of half of the earth's radius, the orbital velocity is
a 3/2 v
b √(3/2)v
c √(2/3)v
d (2/3)v The orbital velocity of an artificial satellite in a circular orbit just above the earth's surface is ν. For a satellite orbiting at an altitude of half of the earth's radius, the orbital velocity is
a 3/2 v
b √(3/2)v
c √(2/3)v
d (2/3)v Homework If the ratio of the orbital distance of two planets d1/d2 =2, what is the ratio of gravitational field experienced by these two planets?
Answer in comment box