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:
- 1200 km
- 540 km
- 1600 km
- 1000 km
The Correct Option is C
Approach Solution - 1
To find the height of the point above the Earth's surface where the gravitational potential is \(-5.12 \times 10^7 \, \text{J} / \text{kg}\) and the acceleration due to gravity is \(6.4 \, \text{m/s}^2\), we can use the relationship between gravitational potential and gravitational acceleration.
The gravitational potential \(V\) at a height \(h\) above the Earth's surface is given by:
\(V = -\frac{GM}{R+h}\)
where \(G\) is the gravitational constant, \(M\) is the mass of the Earth, \(R\) is the radius of the Earth, and \(h\) is the height above the Earth's surface.
The acceleration due to gravity \(g'\) at height \(h\) is given by:
\(g' = \frac{GM}{(R+h)^2}\)
We are given:
- \(V = -5.12 \times 10^7 \, \text{J} / \text{kg}\)
- \(g' = 6.4 \, \text{m/s}^2\)
- \(R = 6400 \, \text{km} = 6400 \times 10^3 \, \text{m}\)
From the formula of gravitational potential, we have:
\(-\frac{GM}{R+h} = -5.12 \times 10^7\)
From the formula for gravitational acceleration at height \(h\), we have:
\(\frac{GM}{(R+h)^2} = 6.4\)
Let's divide the first equation by the second to eliminate \(GM\):
\(\frac{R+h}{(R+h)^2} = \frac{5.12 \times 10^7}{6.4}\)
Simplifying gives us:
\(R+h = \frac{5.12 \times 10^7}{6.4} \cdot (R+h)\)
Solving for \(h\), first calculate:
\(x = \frac{5.12 \times 10^7}{6.4} = 8 \times 10^6\)
Now solve for \(h:\)
\(R+h = 8 \times 10^6\)
Substituting \(R = 6400 \times 10^3\) (the radius of the Earth):
\(6400 \times 10^3 + h = 8 \times 10^6\)
\(h = 8 \times 10^6 - 6400 \times 10^3\)
\(h = 1600 \times 10^3\)
Therefore, \(h = 1600 \, \text{km}\), which matches the correct answer of \(1600 \, \text{km}\).
Approach Solution -2
Using the formula for gravitational potential and gravitational field:
\(\frac{-G M_E}{R_E + h} = -5.12 \times 10^7\) and \(\frac{G M_E}{(R_E + h)^2} = 6.4\)
\(16 \times 10^5\)
(This is in meters)
To convert, divide by 1000:
\(h = 1600 \, \text{km}\)
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Concepts Used:
Units and Measurement
Unit:
A unit of a physical quantity is an arbitrarily chosen standard that is broadly acknowledged by the society and in terms of which other quantities of similar nature may be measured.
Measurement:
The process of measurement is basically a comparison process. To measure a physical quantity, we have to find out how many times a standard amount of that physical quantity is present in the quantity being measured. The number thus obtained is known as the magnitude and the standard chosen is called the unit of the physical quantity.
Read More: Fundamental and Derived Units of Measurement
System of Units:
- CGS system
- FPS system
- MKS system
- SI units
Types of Units:
Fundamental Units -
The units defined for the fundamental quantities are called fundamental units.
Derived Units -
The units of all other physical quantities which are derived from the fundamental units are called the derived units.