Question

The gravitational potential at a point above the surface of earth is \( -5.12 \times 10^7 \, \text{J} / \text{kg} \) and the acceleration due to gravity at that point is \( 6.4 \, \text{m/s}^2 \). Assume that the mean radius of earth to be \( 6400 \, \text{km} \). The height of this point above the earth’s surface is:

• A. 1200 km
• B. 540 km
• C. 1600 km
• D. 1000 km

Answer 1

The Correct Option is C

To determine the altitude above Earth's surface where the gravitational potential is \(-5.12 \times 10^7 \, \text{J} / \text{kg}\) and the gravitational acceleration is \(6.4 \, \text{m/s}^2\), we utilize the relationship between these two quantities.

The gravitational potential \(V\) at an altitude \(h\) from Earth's surface is expressed as:

\(V = -\frac{GM}{R+h}\)

Here, \(G\) represents the gravitational constant,  \(M\) is Earth's mass,  \(R\) is Earth's radius, and \(h\) is the altitude above the surface.

The gravitational acceleration \(g'\) at altitude \(h\) is given by:

\(g' = \frac{GM}{(R+h)^2}\)

Given values are:

• \(V = -5.12 \times 10^7 \, \text{J} / \text{kg}\)
• \(g' = 6.4 \, \text{m/s}^2\)
• \(R = 6400 \, \text{km} = 6400 \times 10^3 \, \text{m}\)

From the gravitational potential formula:

\(-\frac{GM}{R+h} = -5.12 \times 10^7\)

From the gravitational acceleration formula at altitude \(h\):

\(\frac{GM}{(R+h)^2} = 6.4\)

Dividing the first equation by the second eliminates \(GM\):

\(\frac{R+h}{(R+h)^2} = \frac{5.12 \times 10^7}{6.4}\)

Simplification yields:

\(\frac{1}{R+h} = \frac{5.12 \times 10^7}{6.4}\)

Solving for \(h\). First, calculate the value of the right side:

\(\frac{5.12 \times 10^7}{6.4} = 8 \times 10^6\)

Thus:

\(\frac{1}{R+h} = 8 \times 10^6\)

This implies:

\(R+h = \frac{1}{8 \times 10^6}\)

Substituting \(R = 6400 \times 10^3\) (Earth's radius):

\(6400 \times 10^3 + h = \frac{1}{8 \times 10^6}\)

This approach leads to a contradiction. Let's re-evaluate the division step.

Dividing the equation for potential by the equation for acceleration:

\(\frac{V}{g'} = \frac{-GM/(R+h)}{GM/(R+h)^2} = -(R+h)\)

Therefore:

\(R+h = -\frac{V}{g'}\)

Substituting the given values:

\(R+h = -\frac{-5.12 \times 10^7 \, \text{J} / \text{kg}}{6.4 \, \text{m/s}^2}\)

\(R+h = \frac{5.12 \times 10^7}{6.4}\)

\(R+h = 8 \times 10^6 \, \text{m}\)

Substituting the value of Earth's radius, \(R = 6400 \times 10^3 \, \text{m}\):

\(6400 \times 10^3 \, \text{m} + h = 8 \times 10^6 \, \text{m}\)

\(h = 8 \times 10^6 \, \text{m} - 6400 \times 10^3 \, \text{m}\)

\(h = 8000 \times 10^3 \, \text{m} - 6400 \times 10^3 \, \text{m}\)

\(h = 1600 \times 10^3 \, \text{m}\)

Thus, the altitude is \(h = 1600 \, \text{km}\).