The velocity of the satellite at the point C.

Question

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Answer 1

Question

Chapter 5, Problem 90P

(a)

To determine

The velocity of the satellite at the point C.

(a)

Expert Solution

Answer to Problem 90P

The velocity of the satellite at the point C is $3.07 km/s$ .

Explanation of Solution

Write the expression for velocity of the satellite.

$v=2π(r+h)T$ (I)

Here, $v$ is the velocity of the satellite, $r$ is the radius of the Earth, $h$ is the altitude of the satellite above the Earth’s surface, and $T$ is the orbital period of the Earth.

Conclusion:

Substitute $6371 km$ for $r$ , $35800 km$ for $h$ , and $86400 s$ for $T$ in equation (I) to find $v$ .

$v=2π(6371 km+35800 km)86400 s=2(3.14)(6371 km+35800 km)86400 s=3.07 km/s$

Therefore, the velocity of the satellite at the point C is $3.07 km/s$ .

(b)

To determine

The average velocity of the satellite for one quarter of an orbit starts at point A to the end point B.

(b)

Expert Solution

Answer to Problem 90P

The average velocity of the satellite for one quarter of an orbit starts at point A to the end point B is $2.76 km/s$ .

Explanation of Solution

Write the expression for average velocity of the satellite.

$vavg=ΔrΔt$ (II)

Here, $vavg$ is the average velocity, $Δr$ is the change in position, and $Δt$ is the time period.

Substitute $(r+h)2$ for $Δr$ and $T/4$ for $Δt$ in equation (II).

$vavg=(r+h)2T/4=4(r+h)2T$

Conclusion:

Substitute $6371 km$ for $r$ , $35800 km$ for $h$ , and $86400 s$ for $T$ in above relation to find $vavg$ .

$vavg=4(6371 km+35800 km)2(86400 s)=2.76 km/s$

Therefore, the angle does the wheel rotate during the third $1.0 s$ time interval is $450°$ .

(c)

To determine

The average acceleration of the satellite for one quarter of an orbit starts at point A to the end point B.

(c)

Expert Solution

Answer to Problem 90P

The average acceleration of the satellite for one quarter of an orbit starts at point A to the end point B is $0.201 m/s2$ .

Explanation of Solution

Write the expression for average acceleration of the satellite.

$aavg=ΔvΔt$ (III)

Here, $aavg$ is the average acceleration, $Δv$ is the change in velocity, and $Δt$ is the time period.

Since, the average acceleration is in the same direction as,

$Δv=vB−vA=(Δvx)2+(Δvy)2=v2+v2=v2$

Substitute $v2$ for $Δv$ and $T/4$ for $Δt$ in equation (III).

$aavg=v2T/4=4v2T$

Conclusion:

Substitute $3.07 km/s$ for $v$ and $86400 s$ for $T$ in above relation to find $aavg$ .

$aavg=4(3.07 km/s×103 m1 km)2(86400 s)=4(3.07×103 m/s)2(86400 s)=0.201 m/s2$

Therefore, the average acceleration of the satellite for one quarter of an orbit starts at point A to the end point B is $0.201 m/s2$ .

(d)

To determine

The acceleration of the satellite at point D.

(d)

Expert Solution

Answer to Problem 90P

The acceleration of the satellite at point D is $0.224 m/s2$ .

Explanation of Solution

Write the expression for gravitational force.

$F=GmMEr2$ (IV)

Here, $F$ is the gravitational force exerted on the satellite by the Earth, $G$ is the gravitational constant, $m$ is the mass of the satellite, $ME$ is the mass of the Earth, and $r$ is the distance between the satellite and Earth.

Write the expression from Newton’s second law.

$F=mac$ (V)

Here, $F$ is the force exerted on the satellite and $ac$ is the centripetal acceleration.

Conclusion:

Solve the equation (IV) and (V) to find centripetal acceleration.

$mac=GmMEr2ac=GME(r+h)2$

Substitute $6.673×10−11 N⋅m2/kg2$ for $G$ , $5.97×1024 kg$ for $ME$ , $6371 km$ for $r$ , and $35800 km$ for $h$ in above relation to find $ac$ .

$ac=(6.673×10−11 N⋅m2/kg2)(5.97×1024 kg)(6371 km×103 m1 km+35800 km×103 m1 km)2=(6.673×10−11 N⋅m2/kg2)(5.97×1024 kg)(6371×103 m+35800×103 m)2=0.224 m/s2$

Therefore, the acceleration of the satellite at point D is $0.224 m/s2$ .

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Answer 2

Question

Chapter 5, Problem 90P

(a)

To determine

The velocity of the satellite at the point C.

(a)

Expert Solution

Answer to Problem 90P

The velocity of the satellite at the point C is $3.07 km/s$ .

Explanation of Solution

Write the expression for velocity of the satellite.

$v=2π(r+h)T$ (I)

Here, $v$ is the velocity of the satellite, $r$ is the radius of the Earth, $h$ is the altitude of the satellite above the Earth’s surface, and $T$ is the orbital period of the Earth.

Conclusion:

Substitute $6371 km$ for $r$ , $35800 km$ for $h$ , and $86400 s$ for $T$ in equation (I) to find $v$ .

$v=2π(6371 km+35800 km)86400 s=2(3.14)(6371 km+35800 km)86400 s=3.07 km/s$

Therefore, the velocity of the satellite at the point C is $3.07 km/s$ .

(b)

To determine

The average velocity of the satellite for one quarter of an orbit starts at point A to the end point B.

(b)

Expert Solution

Answer to Problem 90P

The average velocity of the satellite for one quarter of an orbit starts at point A to the end point B is $2.76 km/s$ .

Explanation of Solution

Write the expression for average velocity of the satellite.

$vavg=ΔrΔt$ (II)

Here, $vavg$ is the average velocity, $Δr$ is the change in position, and $Δt$ is the time period.

Substitute $(r+h)2$ for $Δr$ and $T/4$ for $Δt$ in equation (II).

$vavg=(r+h)2T/4=4(r+h)2T$

Conclusion:

Substitute $6371 km$ for $r$ , $35800 km$ for $h$ , and $86400 s$ for $T$ in above relation to find $vavg$ .

$vavg=4(6371 km+35800 km)2(86400 s)=2.76 km/s$

Therefore, the angle does the wheel rotate during the third $1.0 s$ time interval is $450°$ .

(c)

To determine

The average acceleration of the satellite for one quarter of an orbit starts at point A to the end point B.

(c)

Expert Solution

Answer to Problem 90P

The average acceleration of the satellite for one quarter of an orbit starts at point A to the end point B is $0.201 m/s2$ .

Explanation of Solution

Write the expression for average acceleration of the satellite.

$aavg=ΔvΔt$ (III)

Here, $aavg$ is the average acceleration, $Δv$ is the change in velocity, and $Δt$ is the time period.

Since, the average acceleration is in the same direction as,

$Δv=vB−vA=(Δvx)2+(Δvy)2=v2+v2=v2$

Substitute $v2$ for $Δv$ and $T/4$ for $Δt$ in equation (III).

$aavg=v2T/4=4v2T$

Conclusion:

Substitute $3.07 km/s$ for $v$ and $86400 s$ for $T$ in above relation to find $aavg$ .

$aavg=4(3.07 km/s×103 m1 km)2(86400 s)=4(3.07×103 m/s)2(86400 s)=0.201 m/s2$

Therefore, the average acceleration of the satellite for one quarter of an orbit starts at point A to the end point B is $0.201 m/s2$ .

(d)

To determine

The acceleration of the satellite at point D.

(d)

Expert Solution

Answer to Problem 90P

The acceleration of the satellite at point D is $0.224 m/s2$ .

Explanation of Solution

Write the expression for gravitational force.

$F=GmMEr2$ (IV)

Here, $F$ is the gravitational force exerted on the satellite by the Earth, $G$ is the gravitational constant, $m$ is the mass of the satellite, $ME$ is the mass of the Earth, and $r$ is the distance between the satellite and Earth.

Write the expression from Newton’s second law.

$F=mac$ (V)

Here, $F$ is the force exerted on the satellite and $ac$ is the centripetal acceleration.

Conclusion:

Solve the equation (IV) and (V) to find centripetal acceleration.

$mac=GmMEr2ac=GME(r+h)2$

Substitute $6.673×10−11 N⋅m2/kg2$ for $G$ , $5.97×1024 kg$ for $ME$ , $6371 km$ for $r$ , and $35800 km$ for $h$ in above relation to find $ac$ .

$ac=(6.673×10−11 N⋅m2/kg2)(5.97×1024 kg)(6371 km×103 m1 km+35800 km×103 m1 km)2=(6.673×10−11 N⋅m2/kg2)(5.97×1024 kg)(6371×103 m+35800×103 m)2=0.224 m/s2$

Therefore, the acceleration of the satellite at point D is $0.224 m/s2$ .

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Answer 3

Question

Chapter 5, Problem 90P

(a)

To determine

The velocity of the satellite at the point C.

(a)

Expert Solution

Answer to Problem 90P

The velocity of the satellite at the point C is $3.07 km/s$ .

Explanation of Solution

Write the expression for velocity of the satellite.

$v=2π(r+h)T$ (I)

Here, $v$ is the velocity of the satellite, $r$ is the radius of the Earth, $h$ is the altitude of the satellite above the Earth’s surface, and $T$ is the orbital period of the Earth.

Conclusion:

Substitute $6371 km$ for $r$ , $35800 km$ for $h$ , and $86400 s$ for $T$ in equation (I) to find $v$ .

$v=2π(6371 km+35800 km)86400 s=2(3.14)(6371 km+35800 km)86400 s=3.07 km/s$

Therefore, the velocity of the satellite at the point C is $3.07 km/s$ .

(b)

To determine

The average velocity of the satellite for one quarter of an orbit starts at point A to the end point B.

(b)

Expert Solution

Answer to Problem 90P

The average velocity of the satellite for one quarter of an orbit starts at point A to the end point B is $2.76 km/s$ .

Explanation of Solution

Write the expression for average velocity of the satellite.

$vavg=ΔrΔt$ (II)

Here, $vavg$ is the average velocity, $Δr$ is the change in position, and $Δt$ is the time period.

Substitute $(r+h)2$ for $Δr$ and $T/4$ for $Δt$ in equation (II).

$vavg=(r+h)2T/4=4(r+h)2T$

Conclusion:

Substitute $6371 km$ for $r$ , $35800 km$ for $h$ , and $86400 s$ for $T$ in above relation to find $vavg$ .

$vavg=4(6371 km+35800 km)2(86400 s)=2.76 km/s$

Therefore, the angle does the wheel rotate during the third $1.0 s$ time interval is $450°$ .

(c)

To determine

The average acceleration of the satellite for one quarter of an orbit starts at point A to the end point B.

(c)

Expert Solution

Answer to Problem 90P

The average acceleration of the satellite for one quarter of an orbit starts at point A to the end point B is $0.201 m/s2$ .

Explanation of Solution

Write the expression for average acceleration of the satellite.

$aavg=ΔvΔt$ (III)

Here, $aavg$ is the average acceleration, $Δv$ is the change in velocity, and $Δt$ is the time period.

Since, the average acceleration is in the same direction as,

$Δv=vB−vA=(Δvx)2+(Δvy)2=v2+v2=v2$

Substitute $v2$ for $Δv$ and $T/4$ for $Δt$ in equation (III).

$aavg=v2T/4=4v2T$

Conclusion:

Substitute $3.07 km/s$ for $v$ and $86400 s$ for $T$ in above relation to find $aavg$ .

$aavg=4(3.07 km/s×103 m1 km)2(86400 s)=4(3.07×103 m/s)2(86400 s)=0.201 m/s2$

Therefore, the average acceleration of the satellite for one quarter of an orbit starts at point A to the end point B is $0.201 m/s2$ .

(d)

To determine

The acceleration of the satellite at point D.

(d)

Expert Solution

Answer to Problem 90P

The acceleration of the satellite at point D is $0.224 m/s2$ .

Explanation of Solution

Write the expression for gravitational force.

$F=GmMEr2$ (IV)

Here, $F$ is the gravitational force exerted on the satellite by the Earth, $G$ is the gravitational constant, $m$ is the mass of the satellite, $ME$ is the mass of the Earth, and $r$ is the distance between the satellite and Earth.

Write the expression from Newton’s second law.

$F=mac$ (V)

Here, $F$ is the force exerted on the satellite and $ac$ is the centripetal acceleration.

Conclusion:

Solve the equation (IV) and (V) to find centripetal acceleration.

$mac=GmMEr2ac=GME(r+h)2$

Substitute $6.673×10−11 N⋅m2/kg2$ for $G$ , $5.97×1024 kg$ for $ME$ , $6371 km$ for $r$ , and $35800 km$ for $h$ in above relation to find $ac$ .

$ac=(6.673×10−11 N⋅m2/kg2)(5.97×1024 kg)(6371 km×103 m1 km+35800 km×103 m1 km)2=(6.673×10−11 N⋅m2/kg2)(5.97×1024 kg)(6371×103 m+35800×103 m)2=0.224 m/s2$

Therefore, the acceleration of the satellite at point D is $0.224 m/s2$ .

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Answer 4

Question

Chapter 5, Problem 90P

(a)

To determine

The velocity of the satellite at the point C.

(a)

Expert Solution

Answer to Problem 90P

The velocity of the satellite at the point C is $3.07 km/s$ .

Explanation of Solution

Write the expression for velocity of the satellite.

$v=2π(r+h)T$ (I)

Here, $v$ is the velocity of the satellite, $r$ is the radius of the Earth, $h$ is the altitude of the satellite above the Earth’s surface, and $T$ is the orbital period of the Earth.

Conclusion:

Substitute $6371 km$ for $r$ , $35800 km$ for $h$ , and $86400 s$ for $T$ in equation (I) to find $v$ .

$v=2π(6371 km+35800 km)86400 s=2(3.14)(6371 km+35800 km)86400 s=3.07 km/s$

Therefore, the velocity of the satellite at the point C is $3.07 km/s$ .

(b)

To determine

The average velocity of the satellite for one quarter of an orbit starts at point A to the end point B.

(b)

Expert Solution

Answer to Problem 90P

The average velocity of the satellite for one quarter of an orbit starts at point A to the end point B is $2.76 km/s$ .

Explanation of Solution

Write the expression for average velocity of the satellite.

$vavg=ΔrΔt$ (II)

Here, $vavg$ is the average velocity, $Δr$ is the change in position, and $Δt$ is the time period.

Substitute $(r+h)2$ for $Δr$ and $T/4$ for $Δt$ in equation (II).

$vavg=(r+h)2T/4=4(r+h)2T$

Conclusion:

Substitute $6371 km$ for $r$ , $35800 km$ for $h$ , and $86400 s$ for $T$ in above relation to find $vavg$ .

$vavg=4(6371 km+35800 km)2(86400 s)=2.76 km/s$

Therefore, the angle does the wheel rotate during the third $1.0 s$ time interval is $450°$ .

(c)

To determine

The average acceleration of the satellite for one quarter of an orbit starts at point A to the end point B.

(c)

Expert Solution

Answer to Problem 90P

The average acceleration of the satellite for one quarter of an orbit starts at point A to the end point B is $0.201 m/s2$ .

Explanation of Solution

Write the expression for average acceleration of the satellite.

$aavg=ΔvΔt$ (III)

Here, $aavg$ is the average acceleration, $Δv$ is the change in velocity, and $Δt$ is the time period.

Since, the average acceleration is in the same direction as,

$Δv=vB−vA=(Δvx)2+(Δvy)2=v2+v2=v2$

Substitute $v2$ for $Δv$ and $T/4$ for $Δt$ in equation (III).

$aavg=v2T/4=4v2T$

Conclusion:

Substitute $3.07 km/s$ for $v$ and $86400 s$ for $T$ in above relation to find $aavg$ .

$aavg=4(3.07 km/s×103 m1 km)2(86400 s)=4(3.07×103 m/s)2(86400 s)=0.201 m/s2$

Therefore, the average acceleration of the satellite for one quarter of an orbit starts at point A to the end point B is $0.201 m/s2$ .

(d)

To determine

The acceleration of the satellite at point D.

(d)

Expert Solution

Answer to Problem 90P

The acceleration of the satellite at point D is $0.224 m/s2$ .

Explanation of Solution

Write the expression for gravitational force.

$F=GmMEr2$ (IV)

Here, $F$ is the gravitational force exerted on the satellite by the Earth, $G$ is the gravitational constant, $m$ is the mass of the satellite, $ME$ is the mass of the Earth, and $r$ is the distance between the satellite and Earth.

Write the expression from Newton’s second law.

$F=mac$ (V)

Here, $F$ is the force exerted on the satellite and $ac$ is the centripetal acceleration.

Conclusion:

Solve the equation (IV) and (V) to find centripetal acceleration.

$mac=GmMEr2ac=GME(r+h)2$

Substitute $6.673×10−11 N⋅m2/kg2$ for $G$ , $5.97×1024 kg$ for $ME$ , $6371 km$ for $r$ , and $35800 km$ for $h$ in above relation to find $ac$ .

$ac=(6.673×10−11 N⋅m2/kg2)(5.97×1024 kg)(6371 km×103 m1 km+35800 km×103 m1 km)2=(6.673×10−11 N⋅m2/kg2)(5.97×1024 kg)(6371×103 m+35800×103 m)2=0.224 m/s2$

Therefore, the acceleration of the satellite at point D is $0.224 m/s2$ .

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Answer 5

Chapter 5, Problem 90P

(a)

To determine

The velocity of the satellite at the point C.

(a)

Expert Solution

Answer to Problem 90P

The velocity of the satellite at the point C is $3.07 km/s$ .

Explanation of Solution

Write the expression for velocity of the satellite.

$v=2π(r+h)T$ (I)

Here, $v$ is the velocity of the satellite, $r$ is the radius of the Earth, $h$ is the altitude of the satellite above the Earth’s surface, and $T$ is the orbital period of the Earth.

Conclusion:

Substitute $6371 km$ for $r$ , $35800 km$ for $h$ , and $86400 s$ for $T$ in equation (I) to find $v$ .

$v=2π(6371 km+35800 km)86400 s=2(3.14)(6371 km+35800 km)86400 s=3.07 km/s$

Therefore, the velocity of the satellite at the point C is $3.07 km/s$ .

(b)

To determine

The average velocity of the satellite for one quarter of an orbit starts at point A to the end point B.

(b)

Expert Solution

Answer to Problem 90P

The average velocity of the satellite for one quarter of an orbit starts at point A to the end point B is $2.76 km/s$ .

Explanation of Solution

Write the expression for average velocity of the satellite.

$vavg=ΔrΔt$ (II)

Here, $vavg$ is the average velocity, $Δr$ is the change in position, and $Δt$ is the time period.

Substitute $(r+h)2$ for $Δr$ and $T/4$ for $Δt$ in equation (II).

$vavg=(r+h)2T/4=4(r+h)2T$

Conclusion:

Substitute $6371 km$ for $r$ , $35800 km$ for $h$ , and $86400 s$ for $T$ in above relation to find $vavg$ .

$vavg=4(6371 km+35800 km)2(86400 s)=2.76 km/s$

Therefore, the angle does the wheel rotate during the third $1.0 s$ time interval is $450°$ .

(c)

To determine

The average acceleration of the satellite for one quarter of an orbit starts at point A to the end point B.

(c)

Expert Solution

Answer to Problem 90P

The average acceleration of the satellite for one quarter of an orbit starts at point A to the end point B is $0.201 m/s2$ .

Explanation of Solution

Write the expression for average acceleration of the satellite.

$aavg=ΔvΔt$ (III)

Here, $aavg$ is the average acceleration, $Δv$ is the change in velocity, and $Δt$ is the time period.

Since, the average acceleration is in the same direction as,

$Δv=vB−vA=(Δvx)2+(Δvy)2=v2+v2=v2$

Substitute $v2$ for $Δv$ and $T/4$ for $Δt$ in equation (III).

$aavg=v2T/4=4v2T$

Conclusion:

Substitute $3.07 km/s$ for $v$ and $86400 s$ for $T$ in above relation to find $aavg$ .

$aavg=4(3.07 km/s×103 m1 km)2(86400 s)=4(3.07×103 m/s)2(86400 s)=0.201 m/s2$

Therefore, the average acceleration of the satellite for one quarter of an orbit starts at point A to the end point B is $0.201 m/s2$ .

(d)

To determine

The acceleration of the satellite at point D.

(d)

Expert Solution

Answer to Problem 90P

The acceleration of the satellite at point D is $0.224 m/s2$ .

Explanation of Solution

Write the expression for gravitational force.

$F=GmMEr2$ (IV)

Here, $F$ is the gravitational force exerted on the satellite by the Earth, $G$ is the gravitational constant, $m$ is the mass of the satellite, $ME$ is the mass of the Earth, and $r$ is the distance between the satellite and Earth.

Write the expression from Newton’s second law.

$F=mac$ (V)

Here, $F$ is the force exerted on the satellite and $ac$ is the centripetal acceleration.

Conclusion:

Solve the equation (IV) and (V) to find centripetal acceleration.

$mac=GmMEr2ac=GME(r+h)2$

Substitute $6.673×10−11 N⋅m2/kg2$ for $G$ , $5.97×1024 kg$ for $ME$ , $6371 km$ for $r$ , and $35800 km$ for $h$ in above relation to find $ac$ .

$ac=(6.673×10−11 N⋅m2/kg2)(5.97×1024 kg)(6371 km×103 m1 km+35800 km×103 m1 km)2=(6.673×10−11 N⋅m2/kg2)(5.97×1024 kg)(6371×103 m+35800×103 m)2=0.224 m/s2$

Therefore, the acceleration of the satellite at point D is $0.224 m/s2$ .

Answer 6

Chapter 5, Problem 90P

(a)

To determine

The velocity of the satellite at the point C.

(a)

Expert Solution

Answer to Problem 90P

The velocity of the satellite at the point C is $3.07 km/s$ .

Explanation of Solution

Write the expression for velocity of the satellite.

$v=2π(r+h)T$ (I)

Here, $v$ is the velocity of the satellite, $r$ is the radius of the Earth, $h$ is the altitude of the satellite above the Earth’s surface, and $T$ is the orbital period of the Earth.

Conclusion:

Substitute $6371 km$ for $r$ , $35800 km$ for $h$ , and $86400 s$ for $T$ in equation (I) to find $v$ .

$v=2π(6371 km+35800 km)86400 s=2(3.14)(6371 km+35800 km)86400 s=3.07 km/s$

Therefore, the velocity of the satellite at the point C is $3.07 km/s$ .

Answer 7

Chapter 5, Problem 90P

(a)

To determine

The velocity of the satellite at the point C.

Answer 8

(a)

Expert Solution

Answer to Problem 90P

The velocity of the satellite at the point C is $3.07 km/s$ .

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

Explanation of Solution

Write the expression for velocity of the satellite.

$v=2π(r+h)T$ (I)

Here, $v$ is the velocity of the satellite, $r$ is the radius of the Earth, $h$ is the altitude of the satellite above the Earth’s surface, and $T$ is the orbital period of the Earth.

Conclusion:

Substitute $6371 km$ for $r$ , $35800 km$ for $h$ , and $86400 s$ for $T$ in equation (I) to find $v$ .

$v=2π(6371 km+35800 km)86400 s=2(3.14)(6371 km+35800 km)86400 s=3.07 km/s$

Therefore, the velocity of the satellite at the point C is $3.07 km/s$ .

Answer 13

(b)

To determine

The average velocity of the satellite for one quarter of an orbit starts at point A to the end point B.

(b)

Expert Solution

Answer to Problem 90P

The average velocity of the satellite for one quarter of an orbit starts at point A to the end point B is $2.76 km/s$ .

Explanation of Solution

Write the expression for average velocity of the satellite.

$vavg=ΔrΔt$ (II)

Here, $vavg$ is the average velocity, $Δr$ is the change in position, and $Δt$ is the time period.

Substitute $(r+h)2$ for $Δr$ and $T/4$ for $Δt$ in equation (II).

$vavg=(r+h)2T/4=4(r+h)2T$

Conclusion:

Substitute $6371 km$ for $r$ , $35800 km$ for $h$ , and $86400 s$ for $T$ in above relation to find $vavg$ .

$vavg=4(6371 km+35800 km)2(86400 s)=2.76 km/s$

Therefore, the angle does the wheel rotate during the third $1.0 s$ time interval is $450°$ .

Answer 14

(b)

To determine

The average velocity of the satellite for one quarter of an orbit starts at point A to the end point B.

Answer 15

(b)

Expert Solution

Answer to Problem 90P

The average velocity of the satellite for one quarter of an orbit starts at point A to the end point B is $2.76 km/s$ .

Answer 16

(b)

Expert Solution

Answer 17

(b)

Expert Solution

Answer 18

Expert Solution

Answer 19

Explanation of Solution

Write the expression for average velocity of the satellite.

$vavg=ΔrΔt$ (II)

Here, $vavg$ is the average velocity, $Δr$ is the change in position, and $Δt$ is the time period.

Substitute $(r+h)2$ for $Δr$ and $T/4$ for $Δt$ in equation (II).

$vavg=(r+h)2T/4=4(r+h)2T$

Conclusion:

Substitute $6371 km$ for $r$ , $35800 km$ for $h$ , and $86400 s$ for $T$ in above relation to find $vavg$ .

$vavg=4(6371 km+35800 km)2(86400 s)=2.76 km/s$

Therefore, the angle does the wheel rotate during the third $1.0 s$ time interval is $450°$ .

Answer 20

(c)

To determine

The average acceleration of the satellite for one quarter of an orbit starts at point A to the end point B.

(c)

Expert Solution

Answer to Problem 90P

The average acceleration of the satellite for one quarter of an orbit starts at point A to the end point B is $0.201 m/s2$ .

Explanation of Solution

Write the expression for average acceleration of the satellite.

$aavg=ΔvΔt$ (III)

Here, $aavg$ is the average acceleration, $Δv$ is the change in velocity, and $Δt$ is the time period.

Since, the average acceleration is in the same direction as,

$Δv=vB−vA=(Δvx)2+(Δvy)2=v2+v2=v2$

Substitute $v2$ for $Δv$ and $T/4$ for $Δt$ in equation (III).

$aavg=v2T/4=4v2T$

Conclusion:

Substitute $3.07 km/s$ for $v$ and $86400 s$ for $T$ in above relation to find $aavg$ .

$aavg=4(3.07 km/s×103 m1 km)2(86400 s)=4(3.07×103 m/s)2(86400 s)=0.201 m/s2$

Therefore, the average acceleration of the satellite for one quarter of an orbit starts at point A to the end point B is $0.201 m/s2$ .

Answer 21

(c)

To determine

The average acceleration of the satellite for one quarter of an orbit starts at point A to the end point B.

Answer 22

(c)

Expert Solution

Answer to Problem 90P

The average acceleration of the satellite for one quarter of an orbit starts at point A to the end point B is $0.201 m/s2$ .

Answer 23

(c)

Expert Solution

Answer 24

(c)

Expert Solution

Answer 25

Expert Solution

Answer 26

Explanation of Solution

Write the expression for average acceleration of the satellite.

$aavg=ΔvΔt$ (III)

Here, $aavg$ is the average acceleration, $Δv$ is the change in velocity, and $Δt$ is the time period.

Since, the average acceleration is in the same direction as,

$Δv=vB−vA=(Δvx)2+(Δvy)2=v2+v2=v2$

Substitute $v2$ for $Δv$ and $T/4$ for $Δt$ in equation (III).

$aavg=v2T/4=4v2T$

Conclusion:

Substitute $3.07 km/s$ for $v$ and $86400 s$ for $T$ in above relation to find $aavg$ .

$aavg=4(3.07 km/s×103 m1 km)2(86400 s)=4(3.07×103 m/s)2(86400 s)=0.201 m/s2$

Therefore, the average acceleration of the satellite for one quarter of an orbit starts at point A to the end point B is $0.201 m/s2$ .

Answer 27

(d)

To determine

The acceleration of the satellite at point D.

(d)

Expert Solution

Answer to Problem 90P

The acceleration of the satellite at point D is $0.224 m/s2$ .

Explanation of Solution

Write the expression for gravitational force.

$F=GmMEr2$ (IV)

Here, $F$ is the gravitational force exerted on the satellite by the Earth, $G$ is the gravitational constant, $m$ is the mass of the satellite, $ME$ is the mass of the Earth, and $r$ is the distance between the satellite and Earth.

Write the expression from Newton’s second law.

$F=mac$ (V)

Here, $F$ is the force exerted on the satellite and $ac$ is the centripetal acceleration.

Conclusion:

Solve the equation (IV) and (V) to find centripetal acceleration.

$mac=GmMEr2ac=GME(r+h)2$

Substitute $6.673×10−11 N⋅m2/kg2$ for $G$ , $5.97×1024 kg$ for $ME$ , $6371 km$ for $r$ , and $35800 km$ for $h$ in above relation to find $ac$ .

$ac=(6.673×10−11 N⋅m2/kg2)(5.97×1024 kg)(6371 km×103 m1 km+35800 km×103 m1 km)2=(6.673×10−11 N⋅m2/kg2)(5.97×1024 kg)(6371×103 m+35800×103 m)2=0.224 m/s2$

Therefore, the acceleration of the satellite at point D is $0.224 m/s2$ .

Answer 28

(d)

To determine

The acceleration of the satellite at point D.

Answer 29

(d)

Expert Solution

Answer to Problem 90P

The acceleration of the satellite at point D is $0.224 m/s2$ .

Answer 30

(d)

Expert Solution

Answer 31

(d)

Expert Solution

Answer 32

Expert Solution

Answer 33

Explanation of Solution

Write the expression for gravitational force.

$F=GmMEr2$ (IV)

Here, $F$ is the gravitational force exerted on the satellite by the Earth, $G$ is the gravitational constant, $m$ is the mass of the satellite, $ME$ is the mass of the Earth, and $r$ is the distance between the satellite and Earth.

Write the expression from Newton’s second law.

$F=mac$ (V)

Here, $F$ is the force exerted on the satellite and $ac$ is the centripetal acceleration.

Conclusion:

Solve the equation (IV) and (V) to find centripetal acceleration.

$mac=GmMEr2ac=GME(r+h)2$

Substitute $6.673×10−11 N⋅m2/kg2$ for $G$ , $5.97×1024 kg$ for $ME$ , $6371 km$ for $r$ , and $35800 km$ for $h$ in above relation to find $ac$ .

$ac=(6.673×10−11 N⋅m2/kg2)(5.97×1024 kg)(6371 km×103 m1 km+35800 km×103 m1 km)2=(6.673×10−11 N⋅m2/kg2)(5.97×1024 kg)(6371×103 m+35800×103 m)2=0.224 m/s2$