2 vertical upward Motion Qs~~~~~Plz (1 Viewer)

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1.the acceleration due to gravity at a point outside the earth is inversely proportional to the square of the distance x from the centre of the earth.
neglecting air resistance, show that if a particle is projected vertically upwards with speed u from a point on the earth's surface, its speed v inany position x is given by

where R is the radius of the Earth and g the acceleration due to gravity at the Earth's surface.
show that the greatest height(above the Earth's surface) H is given by

and prove that if the speed of projection exceeds 12 km/s ,the particle will escape Earth's influence. R=6400km


2. at a point distant r (r >= R) from the centre of the earth , the acceleration due to gravity is proportional to

and is directed towards the earth's centre, also, neglect forces due to all causes other than the earth's gravity.
a body is projected vertically upwards from the surface of the earth with initial speed V.

1)prove that it will escape from the earth if and only if

2)if

prove that the time taken to rise to a height R above the earth's surface is

thx

 

[Trebla]

(1)
a = k/x² (where k is a constant)
v dv/dx = k/x²
∫v dv = ∫(k/x²) dx
v²/2 = - k/x + c
at x = R, v = u, hence c = u²/2 + k/R
v²/2 = - k/x + u²/2 + k/R
.: v² = u² - 2k/x + 2k/R

but when x = R, a = g (where g is acceleration due to gravity)
since: a = k/x²
.: g = k/R²
.: k = gR²
sub into v²:
v² = u² - 2gR²/x + 2gR²/R
.: v² = u² - 2gR²(1/R - 1/x)

Greatest height occurs when its derivative is 0, i.e. dx/dt = v = 0
sub v = 0 and x = R + H
0 = u² - 2gR²(1/R - 1/(R+H))
u² = 2gR²(1/R - 1/(R+H))
u² = 2gR + 2gR²/(R+H)
u² - 2gR = 2gR²/(R+H)
R + H = 2gR²/(u² - 2gR)
H = 2gR²/(u² - 2gR) - R
H = [2gR² - R(u² - 2gR)]/(u² - 2gR)
H = - u²R/(u² - 2gR)
.: H = u²R/(2gR - u²)

From v² = u² - 2gR²(1/R - 1/x)
the particle escapes the Earth's influence when x --> ∞
As x --> ∞, v² --> u² - 2gR
But v² ≥ 0 always
i.e. u² - 2gR ≥ 0
u² ≥ 2gR
.: u ≥ √2gR (u > 0)
sub g = 9.8 m/s² and R = 6400 000 m, into √2gR and you should get 11200 m/s or 11.2 km/s
.: if u exceeds this limit, (e.g. greater than 12km) then it will escape the Earth's influence


(2)

1) similar to previous question with u replaced by V and x replaced by r
2) v² = V² - 2gR²(1/R - 1/r)
sub V = √2gR
v² = 2gR - 2gR + 2gR²/r
v² = 2gR²/r
.: v = R√(2g)/r) (v > 0 and R > 0)
v = dr/dt = R√(2g/r)
Using definite integrals: (where T is time taken to rise height R)
R√(2g) (from 0 to T)∫dt = (from R to 2R)∫√(r) dr
Simplify to make T the subject hence:
.: T = (1/3)(4 - √2)√(R/g)

 

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回复: Re: 2 vertical upward Motion Qs~~~~~Plz

thx
u help me a lot
thx:wave: