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Escape velocity on the surface of earth is $11.2 kms^{–1}$ Escape velocity from a planet whose masse is the same as that of earth and radius 1/4 that of earth is =……….$ kms^{–1}$
$ Ve \alpha {1 \over \sqrt R } $ if R becomes 1/4 then Ve will be 2 times
The velocity with which a projectile must be fired so that it escapes earth’s gravitational does not depend on ..................
Escape velocity does not depends upon the angle of projection
The escape velocity for a body projected vertically upwards from the surface of earth is $11 kms^{-1} $. If the body is projected at an angle of 450 with the vertical, the escape velocity will be ............$kms^{-1}$
The escape velocity is independent of the direction in which the body is projected. Therefore, even if the body is projected at an angle of 45 degrees with the vertical, the escape velocity remains the same, which is $11 kms^{-1}$. The angle of projection does not affect the magnitude of the escape velocity.
The acceleration due to gravity on a planet is same as that on earth and its radius is four times that of earth. What will be the value of escape velocity on that planet if it is Ve on the earth
$ \upsilon = \sqrt {2gR } \Rightarrow {V_p \over V_e } = \sqrt { { g_p \over g_e } . {Rp \over Re} } = \sqrt { 1 \times 4 } = 2 $ Ve = 2 Ve
A particle of mass 10g is kept on the surface of a uniform sphere of mass 100 kg and radiius 10 cm. Find the work to be done aginst the gravitational force between them to take the particle is away from the sphere $(G = 6.67 \times 10^{-11} SI unit)$
potential energy of system of two mass $ U = - { GMm \over R } = { -6.67 \times 10^{-11} \times 100 \times 10 \times 10^{-3 } \over 10 \times 10^{-2} }= 6.67 \times 10^{-10 } J $ so, the amount of work done to take the particle up to infinte will be $ 6.67 \times 10^{-10} J $
A particle of mass M is situated at the center of a spherical shell of same mass and radius a the magnitude of gravitational potential at a point situated at a / 2 distance from the center will be
Vp = Vsphere + Vpartical $ = { GM \over a } + { GM \over a / 2 } = { 3GM \over a } $
The mass and radius of the sun are $1.99 \times10^{30} kg $ and $ R = 6.96 \times 10^8 m$. The escape velocity of rocket from the sun is = ………..km/sec
$ Ve = \sqrt { 2GM \over R } = \sqrt { 2 \times 6.67 \times 10^{-11} \times 1.99 \times 10^{30} \over 6.98 \times 108 } = 618 kms^{-1} $
The mass of a space ship is 1000 kg. It is to be lauched from earth’s surface out into free space the value of g and R (radius of earth) are $10ms^{-2}$ and 6400 km respectively the required energy for this work will be =………….. J
$ W = 0- \left( -{ - GMm \over R } \right) = { GMm \over R } = gR^2 {m \over R} = mgR$ $ = 1000 \times 10 \times 6400 \times 10^3 = 64 \times 10^9 = 6.4 \times 10^{10 } J $
If r represents the radius of the orbit of a satellite of mass m moving around a planet of mas M, the velocity of the satellite is given by
The velocity ( u) of a satellite in orbit around a planet is given by the formula
$ u = rac{GM}{r} $, where G is the gravitational constant, M is the mass of the planet, and r is the radius of the orbit. Therefore, the correct form is $ u^2 = rac{GM}{r}$. This equation is derived from balancing the gravitational force and the centripetal force acting on the satellite.
Two satellites of mass m1 and m2 (m1 > m2 ) are revolving round the earth in cirular orbits of r1 and r2 ( r1 > r2) respectively. Which of the following statement is true regarding their speeds V1 and V2
$ \upsilon = \sqrt { GM \over r } $ $ it r_1 \gt r_2 then V_1 \lt V_2 $
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