The possible values of charges.

Question

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Answer 1

Question

Chapter 16, Problem 86P

(a)

To determine

The possible values of charges.

(a)

Expert Solution

Answer to Problem 86P

The charge of Sun can be $1.712×1020 C$ and the charge of Earth can be $5.148×1014 C$ .

Explanation of Solution

Write an expression for equilibrium condition for gravitational and electrical force.

$k|qS||qE|r2=GmSmEr2$ (I)

Here, $k$ is the Coulomb’s constant, $qS$ is the charge of Sun, $qE$ is the charge of Earth, $r$ is the distance between Earth and Sun, $G$ is the gravitational constant, $mS$ is the mass of Sun and $mE$ is the mass of Earth.

Rewrite equation (I).

$|qS||qE|=GkmSmE$ (II)

Consider,

$|qS|mS=|qE|mE$ (III)

Rewrite equation (III) to find $|qE|$ .

$|qE|=mEmS|qS|$ (IV)

Rewrite equation (II) to find $|qS|$ .

$|qS|=GkmSmE(1|qE|)$ (V)

Substitute equation (IV) in equation (V).

$|qS|=GkmSmE(1mEmS|qS|)=GkmSmE(mSmE|qS|)$ (VI)

Rewrite equation (VI) to find $|qS|$ .

$|qS|=GkmS$ (VII)

Rewrite equation (III) to find $|qS|$ .

$|qS|=mSmE|qE|$ (VIII)

Rewrite equation (II) to find $|qE|$ .

$|qE|=GkmSmE(1|qS|)$ (IX)

Substitute equation (VIII) in equation (IX).

$|qE|=GkmSmE(1mSmE|qE|)=GkmSmE(mEmS|qE|)$ (X)

Rewrite equation (X) to find $|qE|$ .

$qE=GkmE$ (XI)

Conclusion:

Substitute $1.987×1030 kg$ for $mS$ , $6.674×10−11 N⋅m2/kg2$ for $G$ and $8.988×109 N⋅m2/C2$ for $k$ in equation (VII) to find $|qS|$ .

$|qS|=(6.674×10−11 N⋅m2/kg28.988×109 N⋅m2/C2)(1.987×1030 kg)=(0.74×10−20 C2/kg2)(1.987×1030 kg)=1.712×1020 C$

Substitute $5.974×1024 kg$ for $mE$ , $6.674×10−11 N⋅m2/kg2$ for $G$ and $8.988×109 N⋅m2/C2$ for $k$ in equation (XI) to find $|qE|$ .

$|qE|=(6.674×10−11 N⋅m2/kg28.988×109 N⋅m2/C2)(5.974×1024 kg)=(0.74×10−20 C2/kg2)(5.974×1024 kg)=5.148×1014 C$

Thus, the charge of Sun can be $1.712×1020 C$ and the charge of Earth can be $5.148×1014 C$ . Let the charge of Sun is positive and charge of Earth is negative.

(b)

To determine

If charge imbalance explain Earth’s orbit.

(b)

Expert Solution

Answer to Problem 86P

No, the charge imbalance explain Earth’s orbit.

Explanation of Solution

Consider the charge of electron and proton is different. The net charges on astronomical bodies are the sum of charge of the particles. If the magnitude of the charges of the proton and electron were not exactly equal, astronomical bodies will have a total charge with the same sign.

Like charges will repel each other. Since the charges of the astronomical bodies are similar, the force between them would be repulsive. The force responsible for the Earth’s orbit is attractive. Thus, the charge imbalance explain Earth’s orbit.

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Answer 2

Question

Chapter 16, Problem 86P

(a)

To determine

The possible values of charges.

(a)

Expert Solution

Answer to Problem 86P

The charge of Sun can be $1.712×1020 C$ and the charge of Earth can be $5.148×1014 C$ .

Explanation of Solution

Write an expression for equilibrium condition for gravitational and electrical force.

$k|qS||qE|r2=GmSmEr2$ (I)

Here, $k$ is the Coulomb’s constant, $qS$ is the charge of Sun, $qE$ is the charge of Earth, $r$ is the distance between Earth and Sun, $G$ is the gravitational constant, $mS$ is the mass of Sun and $mE$ is the mass of Earth.

Rewrite equation (I).

$|qS||qE|=GkmSmE$ (II)

Consider,

$|qS|mS=|qE|mE$ (III)

Rewrite equation (III) to find $|qE|$ .

$|qE|=mEmS|qS|$ (IV)

Rewrite equation (II) to find $|qS|$ .

$|qS|=GkmSmE(1|qE|)$ (V)

Substitute equation (IV) in equation (V).

$|qS|=GkmSmE(1mEmS|qS|)=GkmSmE(mSmE|qS|)$ (VI)

Rewrite equation (VI) to find $|qS|$ .

$|qS|=GkmS$ (VII)

Rewrite equation (III) to find $|qS|$ .

$|qS|=mSmE|qE|$ (VIII)

Rewrite equation (II) to find $|qE|$ .

$|qE|=GkmSmE(1|qS|)$ (IX)

Substitute equation (VIII) in equation (IX).

$|qE|=GkmSmE(1mSmE|qE|)=GkmSmE(mEmS|qE|)$ (X)

Rewrite equation (X) to find $|qE|$ .

$qE=GkmE$ (XI)

Conclusion:

Substitute $1.987×1030 kg$ for $mS$ , $6.674×10−11 N⋅m2/kg2$ for $G$ and $8.988×109 N⋅m2/C2$ for $k$ in equation (VII) to find $|qS|$ .

$|qS|=(6.674×10−11 N⋅m2/kg28.988×109 N⋅m2/C2)(1.987×1030 kg)=(0.74×10−20 C2/kg2)(1.987×1030 kg)=1.712×1020 C$

Substitute $5.974×1024 kg$ for $mE$ , $6.674×10−11 N⋅m2/kg2$ for $G$ and $8.988×109 N⋅m2/C2$ for $k$ in equation (XI) to find $|qE|$ .

$|qE|=(6.674×10−11 N⋅m2/kg28.988×109 N⋅m2/C2)(5.974×1024 kg)=(0.74×10−20 C2/kg2)(5.974×1024 kg)=5.148×1014 C$

Thus, the charge of Sun can be $1.712×1020 C$ and the charge of Earth can be $5.148×1014 C$ . Let the charge of Sun is positive and charge of Earth is negative.

(b)

To determine

If charge imbalance explain Earth’s orbit.

(b)

Expert Solution

Answer to Problem 86P

No, the charge imbalance explain Earth’s orbit.

Explanation of Solution

Consider the charge of electron and proton is different. The net charges on astronomical bodies are the sum of charge of the particles. If the magnitude of the charges of the proton and electron were not exactly equal, astronomical bodies will have a total charge with the same sign.

Like charges will repel each other. Since the charges of the astronomical bodies are similar, the force between them would be repulsive. The force responsible for the Earth’s orbit is attractive. Thus, the charge imbalance explain Earth’s orbit.

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Answer 3

Question

Chapter 16, Problem 86P

(a)

To determine

The possible values of charges.

(a)

Expert Solution

Answer to Problem 86P

The charge of Sun can be $1.712×1020 C$ and the charge of Earth can be $5.148×1014 C$ .

Explanation of Solution

Write an expression for equilibrium condition for gravitational and electrical force.

$k|qS||qE|r2=GmSmEr2$ (I)

Here, $k$ is the Coulomb’s constant, $qS$ is the charge of Sun, $qE$ is the charge of Earth, $r$ is the distance between Earth and Sun, $G$ is the gravitational constant, $mS$ is the mass of Sun and $mE$ is the mass of Earth.

Rewrite equation (I).

$|qS||qE|=GkmSmE$ (II)

Consider,

$|qS|mS=|qE|mE$ (III)

Rewrite equation (III) to find $|qE|$ .

$|qE|=mEmS|qS|$ (IV)

Rewrite equation (II) to find $|qS|$ .

$|qS|=GkmSmE(1|qE|)$ (V)

Substitute equation (IV) in equation (V).

$|qS|=GkmSmE(1mEmS|qS|)=GkmSmE(mSmE|qS|)$ (VI)

Rewrite equation (VI) to find $|qS|$ .

$|qS|=GkmS$ (VII)

Rewrite equation (III) to find $|qS|$ .

$|qS|=mSmE|qE|$ (VIII)

Rewrite equation (II) to find $|qE|$ .

$|qE|=GkmSmE(1|qS|)$ (IX)

Substitute equation (VIII) in equation (IX).

$|qE|=GkmSmE(1mSmE|qE|)=GkmSmE(mEmS|qE|)$ (X)

Rewrite equation (X) to find $|qE|$ .

$qE=GkmE$ (XI)

Conclusion:

Substitute $1.987×1030 kg$ for $mS$ , $6.674×10−11 N⋅m2/kg2$ for $G$ and $8.988×109 N⋅m2/C2$ for $k$ in equation (VII) to find $|qS|$ .

$|qS|=(6.674×10−11 N⋅m2/kg28.988×109 N⋅m2/C2)(1.987×1030 kg)=(0.74×10−20 C2/kg2)(1.987×1030 kg)=1.712×1020 C$

Substitute $5.974×1024 kg$ for $mE$ , $6.674×10−11 N⋅m2/kg2$ for $G$ and $8.988×109 N⋅m2/C2$ for $k$ in equation (XI) to find $|qE|$ .

$|qE|=(6.674×10−11 N⋅m2/kg28.988×109 N⋅m2/C2)(5.974×1024 kg)=(0.74×10−20 C2/kg2)(5.974×1024 kg)=5.148×1014 C$

Thus, the charge of Sun can be $1.712×1020 C$ and the charge of Earth can be $5.148×1014 C$ . Let the charge of Sun is positive and charge of Earth is negative.

(b)

To determine

If charge imbalance explain Earth’s orbit.

(b)

Expert Solution

Answer to Problem 86P

No, the charge imbalance explain Earth’s orbit.

Explanation of Solution

Consider the charge of electron and proton is different. The net charges on astronomical bodies are the sum of charge of the particles. If the magnitude of the charges of the proton and electron were not exactly equal, astronomical bodies will have a total charge with the same sign.

Like charges will repel each other. Since the charges of the astronomical bodies are similar, the force between them would be repulsive. The force responsible for the Earth’s orbit is attractive. Thus, the charge imbalance explain Earth’s orbit.

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Answer 4

Question

Chapter 16, Problem 86P

(a)

To determine

The possible values of charges.

(a)

Expert Solution

Answer to Problem 86P

The charge of Sun can be $1.712×1020 C$ and the charge of Earth can be $5.148×1014 C$ .

Explanation of Solution

Write an expression for equilibrium condition for gravitational and electrical force.

$k|qS||qE|r2=GmSmEr2$ (I)

Here, $k$ is the Coulomb’s constant, $qS$ is the charge of Sun, $qE$ is the charge of Earth, $r$ is the distance between Earth and Sun, $G$ is the gravitational constant, $mS$ is the mass of Sun and $mE$ is the mass of Earth.

Rewrite equation (I).

$|qS||qE|=GkmSmE$ (II)

Consider,

$|qS|mS=|qE|mE$ (III)

Rewrite equation (III) to find $|qE|$ .

$|qE|=mEmS|qS|$ (IV)

Rewrite equation (II) to find $|qS|$ .

$|qS|=GkmSmE(1|qE|)$ (V)

Substitute equation (IV) in equation (V).

$|qS|=GkmSmE(1mEmS|qS|)=GkmSmE(mSmE|qS|)$ (VI)

Rewrite equation (VI) to find $|qS|$ .

$|qS|=GkmS$ (VII)

Rewrite equation (III) to find $|qS|$ .

$|qS|=mSmE|qE|$ (VIII)

Rewrite equation (II) to find $|qE|$ .

$|qE|=GkmSmE(1|qS|)$ (IX)

Substitute equation (VIII) in equation (IX).

$|qE|=GkmSmE(1mSmE|qE|)=GkmSmE(mEmS|qE|)$ (X)

Rewrite equation (X) to find $|qE|$ .

$qE=GkmE$ (XI)

Conclusion:

Substitute $1.987×1030 kg$ for $mS$ , $6.674×10−11 N⋅m2/kg2$ for $G$ and $8.988×109 N⋅m2/C2$ for $k$ in equation (VII) to find $|qS|$ .

$|qS|=(6.674×10−11 N⋅m2/kg28.988×109 N⋅m2/C2)(1.987×1030 kg)=(0.74×10−20 C2/kg2)(1.987×1030 kg)=1.712×1020 C$

Substitute $5.974×1024 kg$ for $mE$ , $6.674×10−11 N⋅m2/kg2$ for $G$ and $8.988×109 N⋅m2/C2$ for $k$ in equation (XI) to find $|qE|$ .

$|qE|=(6.674×10−11 N⋅m2/kg28.988×109 N⋅m2/C2)(5.974×1024 kg)=(0.74×10−20 C2/kg2)(5.974×1024 kg)=5.148×1014 C$

Thus, the charge of Sun can be $1.712×1020 C$ and the charge of Earth can be $5.148×1014 C$ . Let the charge of Sun is positive and charge of Earth is negative.

(b)

To determine

If charge imbalance explain Earth’s orbit.

(b)

Expert Solution

Answer to Problem 86P

No, the charge imbalance explain Earth’s orbit.

Explanation of Solution

Consider the charge of electron and proton is different. The net charges on astronomical bodies are the sum of charge of the particles. If the magnitude of the charges of the proton and electron were not exactly equal, astronomical bodies will have a total charge with the same sign.

Like charges will repel each other. Since the charges of the astronomical bodies are similar, the force between them would be repulsive. The force responsible for the Earth’s orbit is attractive. Thus, the charge imbalance explain Earth’s orbit.

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Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 5

Chapter 16, Problem 86P

(a)

To determine

The possible values of charges.

(a)

Expert Solution

Answer to Problem 86P

The charge of Sun can be $1.712×1020 C$ and the charge of Earth can be $5.148×1014 C$ .

Explanation of Solution

Write an expression for equilibrium condition for gravitational and electrical force.

$k|qS||qE|r2=GmSmEr2$ (I)

Here, $k$ is the Coulomb’s constant, $qS$ is the charge of Sun, $qE$ is the charge of Earth, $r$ is the distance between Earth and Sun, $G$ is the gravitational constant, $mS$ is the mass of Sun and $mE$ is the mass of Earth.

Rewrite equation (I).

$|qS||qE|=GkmSmE$ (II)

Consider,

$|qS|mS=|qE|mE$ (III)

Rewrite equation (III) to find $|qE|$ .

$|qE|=mEmS|qS|$ (IV)

Rewrite equation (II) to find $|qS|$ .

$|qS|=GkmSmE(1|qE|)$ (V)

Substitute equation (IV) in equation (V).

$|qS|=GkmSmE(1mEmS|qS|)=GkmSmE(mSmE|qS|)$ (VI)

Rewrite equation (VI) to find $|qS|$ .

$|qS|=GkmS$ (VII)

Rewrite equation (III) to find $|qS|$ .

$|qS|=mSmE|qE|$ (VIII)

Rewrite equation (II) to find $|qE|$ .

$|qE|=GkmSmE(1|qS|)$ (IX)

Substitute equation (VIII) in equation (IX).

$|qE|=GkmSmE(1mSmE|qE|)=GkmSmE(mEmS|qE|)$ (X)

Rewrite equation (X) to find $|qE|$ .

$qE=GkmE$ (XI)

Conclusion:

Substitute $1.987×1030 kg$ for $mS$ , $6.674×10−11 N⋅m2/kg2$ for $G$ and $8.988×109 N⋅m2/C2$ for $k$ in equation (VII) to find $|qS|$ .

$|qS|=(6.674×10−11 N⋅m2/kg28.988×109 N⋅m2/C2)(1.987×1030 kg)=(0.74×10−20 C2/kg2)(1.987×1030 kg)=1.712×1020 C$

Substitute $5.974×1024 kg$ for $mE$ , $6.674×10−11 N⋅m2/kg2$ for $G$ and $8.988×109 N⋅m2/C2$ for $k$ in equation (XI) to find $|qE|$ .

$|qE|=(6.674×10−11 N⋅m2/kg28.988×109 N⋅m2/C2)(5.974×1024 kg)=(0.74×10−20 C2/kg2)(5.974×1024 kg)=5.148×1014 C$

Thus, the charge of Sun can be $1.712×1020 C$ and the charge of Earth can be $5.148×1014 C$ . Let the charge of Sun is positive and charge of Earth is negative.

(b)

To determine

If charge imbalance explain Earth’s orbit.

(b)

Expert Solution

Answer to Problem 86P

No, the charge imbalance explain Earth’s orbit.

Explanation of Solution

Consider the charge of electron and proton is different. The net charges on astronomical bodies are the sum of charge of the particles. If the magnitude of the charges of the proton and electron were not exactly equal, astronomical bodies will have a total charge with the same sign.

Like charges will repel each other. Since the charges of the astronomical bodies are similar, the force between them would be repulsive. The force responsible for the Earth’s orbit is attractive. Thus, the charge imbalance explain Earth’s orbit.

Answer 6

Chapter 16, Problem 86P

(a)

To determine

The possible values of charges.

(a)

Expert Solution

Answer to Problem 86P

The charge of Sun can be $1.712×1020 C$ and the charge of Earth can be $5.148×1014 C$ .

Explanation of Solution

Write an expression for equilibrium condition for gravitational and electrical force.

$k|qS||qE|r2=GmSmEr2$ (I)

Here, $k$ is the Coulomb’s constant, $qS$ is the charge of Sun, $qE$ is the charge of Earth, $r$ is the distance between Earth and Sun, $G$ is the gravitational constant, $mS$ is the mass of Sun and $mE$ is the mass of Earth.

Rewrite equation (I).

$|qS||qE|=GkmSmE$ (II)

Consider,

$|qS|mS=|qE|mE$ (III)

Rewrite equation (III) to find $|qE|$ .

$|qE|=mEmS|qS|$ (IV)

Rewrite equation (II) to find $|qS|$ .

$|qS|=GkmSmE(1|qE|)$ (V)

Substitute equation (IV) in equation (V).

$|qS|=GkmSmE(1mEmS|qS|)=GkmSmE(mSmE|qS|)$ (VI)

Rewrite equation (VI) to find $|qS|$ .

$|qS|=GkmS$ (VII)

Rewrite equation (III) to find $|qS|$ .

$|qS|=mSmE|qE|$ (VIII)

Rewrite equation (II) to find $|qE|$ .

$|qE|=GkmSmE(1|qS|)$ (IX)

Substitute equation (VIII) in equation (IX).

$|qE|=GkmSmE(1mSmE|qE|)=GkmSmE(mEmS|qE|)$ (X)

Rewrite equation (X) to find $|qE|$ .

$qE=GkmE$ (XI)

Conclusion:

Substitute $1.987×1030 kg$ for $mS$ , $6.674×10−11 N⋅m2/kg2$ for $G$ and $8.988×109 N⋅m2/C2$ for $k$ in equation (VII) to find $|qS|$ .

$|qS|=(6.674×10−11 N⋅m2/kg28.988×109 N⋅m2/C2)(1.987×1030 kg)=(0.74×10−20 C2/kg2)(1.987×1030 kg)=1.712×1020 C$

Substitute $5.974×1024 kg$ for $mE$ , $6.674×10−11 N⋅m2/kg2$ for $G$ and $8.988×109 N⋅m2/C2$ for $k$ in equation (XI) to find $|qE|$ .

$|qE|=(6.674×10−11 N⋅m2/kg28.988×109 N⋅m2/C2)(5.974×1024 kg)=(0.74×10−20 C2/kg2)(5.974×1024 kg)=5.148×1014 C$

Thus, the charge of Sun can be $1.712×1020 C$ and the charge of Earth can be $5.148×1014 C$ . Let the charge of Sun is positive and charge of Earth is negative.

Answer 7

Chapter 16, Problem 86P

(a)

To determine

The possible values of charges.

Answer 8

(a)

Expert Solution

Answer to Problem 86P

The charge of Sun can be $1.712×1020 C$ and the charge of Earth can be $5.148×1014 C$ .

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

Explanation of Solution

Write an expression for equilibrium condition for gravitational and electrical force.

$k|qS||qE|r2=GmSmEr2$ (I)

Here, $k$ is the Coulomb’s constant, $qS$ is the charge of Sun, $qE$ is the charge of Earth, $r$ is the distance between Earth and Sun, $G$ is the gravitational constant, $mS$ is the mass of Sun and $mE$ is the mass of Earth.

Rewrite equation (I).

$|qS||qE|=GkmSmE$ (II)

Consider,

$|qS|mS=|qE|mE$ (III)

Rewrite equation (III) to find $|qE|$ .

$|qE|=mEmS|qS|$ (IV)

Rewrite equation (II) to find $|qS|$ .

$|qS|=GkmSmE(1|qE|)$ (V)

Substitute equation (IV) in equation (V).

$|qS|=GkmSmE(1mEmS|qS|)=GkmSmE(mSmE|qS|)$ (VI)

Rewrite equation (VI) to find $|qS|$ .

$|qS|=GkmS$ (VII)

Rewrite equation (III) to find $|qS|$ .

$|qS|=mSmE|qE|$ (VIII)

Rewrite equation (II) to find $|qE|$ .

$|qE|=GkmSmE(1|qS|)$ (IX)

Substitute equation (VIII) in equation (IX).

$|qE|=GkmSmE(1mSmE|qE|)=GkmSmE(mEmS|qE|)$ (X)

Rewrite equation (X) to find $|qE|$ .

$qE=GkmE$ (XI)

Conclusion:

Substitute $1.987×1030 kg$ for $mS$ , $6.674×10−11 N⋅m2/kg2$ for $G$ and $8.988×109 N⋅m2/C2$ for $k$ in equation (VII) to find $|qS|$ .

$|qS|=(6.674×10−11 N⋅m2/kg28.988×109 N⋅m2/C2)(1.987×1030 kg)=(0.74×10−20 C2/kg2)(1.987×1030 kg)=1.712×1020 C$

Substitute $5.974×1024 kg$ for $mE$ , $6.674×10−11 N⋅m2/kg2$ for $G$ and $8.988×109 N⋅m2/C2$ for $k$ in equation (XI) to find $|qE|$ .

$|qE|=(6.674×10−11 N⋅m2/kg28.988×109 N⋅m2/C2)(5.974×1024 kg)=(0.74×10−20 C2/kg2)(5.974×1024 kg)=5.148×1014 C$

Thus, the charge of Sun can be $1.712×1020 C$ and the charge of Earth can be $5.148×1014 C$ . Let the charge of Sun is positive and charge of Earth is negative.

Answer 13

(b)

To determine

If charge imbalance explain Earth’s orbit.

(b)

Expert Solution

Answer to Problem 86P

No, the charge imbalance explain Earth’s orbit.

Explanation of Solution

Consider the charge of electron and proton is different. The net charges on astronomical bodies are the sum of charge of the particles. If the magnitude of the charges of the proton and electron were not exactly equal, astronomical bodies will have a total charge with the same sign.

Like charges will repel each other. Since the charges of the astronomical bodies are similar, the force between them would be repulsive. The force responsible for the Earth’s orbit is attractive. Thus, the charge imbalance explain Earth’s orbit.

Answer 14

(b)

To determine

If charge imbalance explain Earth’s orbit.

Answer 15

(b)

Expert Solution

Answer to Problem 86P

No, the charge imbalance explain Earth’s orbit.

Answer 16

(b)

Expert Solution

Answer 17

(b)

Expert Solution

Answer 18

Expert Solution

Answer 19

Explanation of Solution

Consider the charge of electron and proton is different. The net charges on astronomical bodies are the sum of charge of the particles. If the magnitude of the charges of the proton and electron were not exactly equal, astronomical bodies will have a total charge with the same sign.

Like charges will repel each other. Since the charges of the astronomical bodies are similar, the force between them would be repulsive. The force responsible for the Earth’s orbit is attractive. Thus, the charge imbalance explain Earth’s orbit.