Question

At what height from the surface of earth the gravitation potential and the value of g are $- 5.4 \times 10^7 \, J \, kg^{-2}$ and $6.0 \, ms^{-2}$ respectively ? Take the radius of earth as $6400 \,km$ :

• A. 1600 km
• B. 1400 km
• C. 2000 km
• D. 2600 km

Answer 1

The Correct Option is D

 To determine the height above the Earth's surface where the gravitational potential and gravitational acceleration are given, we need to use the following concepts:

The gravitational potential energy \(V\) at a distance \(r\) from the center of a spherical body like Earth is given by:

\(V = - \frac{G M}{r}\)

where:

• \(G\) is the gravitational constant, approximately \(6.67 \times 10^{-11} \, \text{Nm}^2/\text{kg}^2\)
• \(M\) is the mass of the Earth, about \(5.98 \times 10^{24} \, \text{kg}\)

Given potential \(V = -5.4 \times 10^7 \, J \, kg^{-1}\) and gravitational acceleration \(g = 6.0 \, \text{ms}^{-2}\).

The acceleration due to gravity at height \(h\) is given by:

\(g = \frac{G M}{r^2}\)

Given that \(g = 6.0 \, \text{ms}^{-2}\), equate it to the formula:

\(6.0 = \frac{G M}{r^2}\) (Equation 1)

Now, using the given potential:

\(-5.4 \times 10^7 = - \frac{G M}{r}\) (Equation 2)

From Equation 2, we can solve for \(r\):

\(r = \frac{G M}{5.4 \times 10^7}\)

From Equation 1, we can express \(G M\) in terms of \(r^2\):

\(G M = 6.0 \times r^2\)

Substituting this into the expression for \(r\):

\(r = \frac{6.0 \times r^2}{5.4 \times 10^7}\)

Simplifying this equation gives:

\(r = \frac{6.0}{5.4 \times 10^7} \times r^2\)

Assuming \(r = R + h\), where \(R = 6400 \, \text{km}\) (radius of Earth) and \(h\) is the height above the Earth's surface:

Calculate the value of \(r\) using:

\(\frac{6.0 \times r^2}{5.4 \times 10^7} = r\)

Solving this gives:

\(r = 9000 \, \text{km}\)

Thus, the height \(h\) from the Earth's surface is:

\(h = r - R = 9000 \, \text{km} - 6400 \, \text{km} = 2600 \, \text{km}\)

Therefore, the correct answer is:

 

2600 km