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2) At what distance from Earth's surface is the acceleration due to gravity 7.33m/s^2 (9.97times 10^5m)

Question

2) At what distance from Earth's surface is the acceleration due to gravity 7.33m/s^2 (9.97times 10^5m)

2) At what distance from Earth's surface is the acceleration due to gravity
7.33m/s^2
(9.97times 10^5m)

Solution

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ElliotMaster · Tutor for 5 years

Answer

Here's how to solve this problem:

**1. Understand the concept:**

The acceleration due to gravity (g) decreases as you move further away from the Earth's center. The relationship is given by Newton's Law of Universal Gravitation:

g = (G * M) / r²

Where:

* g is the acceleration due to gravity
* G is the gravitational constant (6.674 x 10⁻¹¹ N⋅m²/kg²)
* M is the mass of the Earth (5.972 x 10²⁴ kg)
* r is the distance from the center of the Earth

**2. Set up the equation:**

We are given g = 7.33 m/s² and we want to find r. We know G and M.

7.33 m/s² = (6.674 x 10⁻¹¹ N⋅m²/kg² * 5.972 x 10²⁴ kg) / r²

**3. Solve for r:**

* Multiply both sides by r²:

7.33 m/s² * r² = (6.674 x 10⁻¹¹ N⋅m²/kg² * 5.972 x 10²⁴ kg)

* Divide both sides by 7.33 m/s²:

r² = (6.674 x 10⁻¹¹ N⋅m²/kg² * 5.972 x 10²⁴ kg) / 7.33 m/s²

r² ≈ 5.456 x 10¹³ m²

* Take the square root of both sides:

r ≈ 7.386 x 10⁶ m

**4. Calculate the distance from Earth's surface:**

The value of 'r' we calculated is the distance from the *center* of the Earth. To find the distance from the *surface*, we need to subtract the Earth's radius (R), which is approximately 6.371 x 10⁶ m.

Distance from surface = r - R

Distance from surface ≈ 7.386 x 10⁶ m - 6.371 x 10⁶ m

Distance from surface ≈ 1.015 x 10⁶ m or 1015 km

**Therefore, the distance from the Earth's surface where the acceleration due to gravity is 7.33 m/s² is approximately 1.015 x 10⁶ m or 1015 km.** The provided answer of 9.97 x 10⁵ m is incorrect.
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