Equation (2-30) is an approximation correct only if the gravitational time-dilation effect is small. In this exercise, it is also assumed to be small but we still allow for a nonuniform gravitational field. We start with (2-29), based on the Doppler effect in the accelerating frame. Consider two elevations, the lower at r 1 and the upper at r 1 + d r . Equation (2-29) becomes f ( r 1 + d r ) f ( r 1 ) = ( 1 − g ( r 1 ) d r c 2 ) Similarly, if we consider elevations r 1 + d r and r 1 + 2 d r , we have f ( r 1 + 2 d r ) f ( r 1 + d r ) = ( 1 − g ( r 1 + d r ) d r c 2 ) We continue the process, incrementing r by dr , until we reach r 2 . f ( r 2 ) f ( r 2 − d r ) = ( 1 − g ( r 2 + d r ) d r c 2 ) Now imagine multiplying the left sides of all the equations and setting the product equal to the product of all the right sides. (a) Argue that the left side of the product is simply f ( r 2 ) / f ( r 1 ) . (b) Assuming that the term g d r / c 2 in each individual equation is very small, so that products of such terms can be ignored, argue that the right side of the product is 1 − 1 c 2 ∫ g ( r ) d r (c) Deduce g ( r ) from Newton’s universal law of gravitation, then argue that equation (2-31) follows from the result, just as (2-30) does from (2-29).
Equation (2-30) is an approximation correct only if the gravitational time-dilation effect is small. In this exercise, it is also assumed to be small but we still allow for a nonuniform gravitational field. We start with (2-29), based on the Doppler effect in the accelerating frame. Consider two elevations, the lower at r 1 and the upper at r 1 + d r . Equation (2-29) becomes f ( r 1 + d r ) f ( r 1 ) = ( 1 − g ( r 1 ) d r c 2 ) Similarly, if we consider elevations r 1 + d r and r 1 + 2 d r , we have f ( r 1 + 2 d r ) f ( r 1 + d r ) = ( 1 − g ( r 1 + d r ) d r c 2 ) We continue the process, incrementing r by dr , until we reach r 2 . f ( r 2 ) f ( r 2 − d r ) = ( 1 − g ( r 2 + d r ) d r c 2 ) Now imagine multiplying the left sides of all the equations and setting the product equal to the product of all the right sides. (a) Argue that the left side of the product is simply f ( r 2 ) / f ( r 1 ) . (b) Assuming that the term g d r / c 2 in each individual equation is very small, so that products of such terms can be ignored, argue that the right side of the product is 1 − 1 c 2 ∫ g ( r ) d r (c) Deduce g ( r ) from Newton’s universal law of gravitation, then argue that equation (2-31) follows from the result, just as (2-30) does from (2-29).
Equation (2-30) is an approximation correct only if the gravitational time-dilation effect is small. In this exercise, it is also assumed to be small but we still allow for a nonuniform gravitational field. We start with (2-29), based on the Doppler effect in the accelerating frame. Consider two elevations, the lower at
r
1
and the upper at
r
1
+
d
r
. Equation (2-29) becomes
f
(
r
1
+
d
r
)
f
(
r
1
)
=
(
1
−
g
(
r
1
)
d
r
c
2
)
Similarly, if we consider elevations
r
1
+
d
r
and
r
1
+
2
d
r
, we have
f
(
r
1
+
2
d
r
)
f
(
r
1
+
d
r
)
=
(
1
−
g
(
r
1
+
d
r
)
d
r
c
2
)
We continue the process, incrementing r by dr, until we reach
r
2
.
f
(
r
2
)
f
(
r
2
−
d
r
)
=
(
1
−
g
(
r
2
+
d
r
)
d
r
c
2
)
Now imagine multiplying the left sides of all the equations and setting the product equal to the product of all the right sides. (a) Argue that the left side of the product is simply
f
(
r
2
)
/
f
(
r
1
)
. (b) Assuming that the term
g
d
r
/
c
2
in each individual equation is very small, so that products of such terms can be ignored, argue that the right side of the product is
1
−
1
c
2
∫
g
(
r
)
d
r
(c) Deduce g(r) from Newton’s universal law of gravitation, then argue that equation (2-31) follows from the result, just as (2-30) does from (2-29).
A planar double pendulum consists of two point masses \[m_1 = 1.00~\mathrm{kg}, \qquad m_2 = 1.00~\mathrm{kg}\]connected by massless, rigid rods of lengths \[L_1 = 1.00~\mathrm{m}, \qquad L_2 = 1.20~\mathrm{m}.\]The upper rod is hinged to a fixed pivot; gravity acts vertically downward with\[g = 9.81~\mathrm{m\,s^{-2}}.\]Define the generalized coordinates \(\theta_1,\theta_2\) as the angles each rod makes with thedownward vertical (positive anticlockwise, measured in radians unless stated otherwise).At \(t=0\) the system is released from rest with \[\theta_1(0)=120^{\circ}, \qquad\theta_2(0)=-10^{\circ}, \qquad\dot{\theta}_1(0)=\dot{\theta}_2(0)=0 .\]Using the exact nonlinear equations of motion (no small-angle or planar-pendulumapproximations) and assuming the rods never stretch or slip, determine the angle\(\theta_2\) at the instant\[t = 10.0~\mathrm{s}.\]Give the result in degrees, in the interval \((-180^{\circ},180^{\circ}]\).
What are the expected readings of the ammeter and voltmeter for the circuit in the figure below? (R = 5.60 Ω, ΔV = 6.30 V)
ammeter
I =
simple diagram to illustrate the setup for each law- coulombs law and biot savart law
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