Question

Find the gravitational potential energy of a body of mass 10 kg when it is at a height of 6400 km from the earth's surface. [Given : mass of the earth and radius of the earth.]

Answer 1

Answer:

$6.27\times 10^8$ J

Explanation:

Given that

Mass of body, m = 10 kg

Height of body, h = 6400 km

Let us convert distance from km to meters.

1 km = 1000 m

6400 km = 6400000 m

So, h = 6400000 m

Acceleration due to gravity, g = 9.8 m/$s^{2}$

Formula for Gravitational Potential Energy is given as:

$PE_G = mgh$

Putting the values of m, g and h:

$PE_G = 10 \times 6400000 \times 9.8\\\Rightarrow PE_G = 6400000 \times 98\\\Rightarrow PE_G = 627200000\\\Rightarrow PE_G = 6.27 \times 10^8\ J$

So, Gravitational Potential Energy of the given body is $6.27\times 10^8$ J.

Answer 2

The gravitational Potential Energy is 3.127 × 10⁸ Joule

Explanation:

Given:

Mass of body = 10 kg

Mass of Earth $=6\times 10^{24}$ kg

Height of the body above the surface of the Earth = 6400 kg

Radius of the Earth = 6400 km

To find out:

Gravitational Potential Energy of the body at the height of 6400 km

Solution:

We know that Potential Energy = mgh

But this formula is applicable only for the bodies that are near the surface of the Earth where we can take g as constant

For bodies sufficiently away from the Earth, the Gravitational Potential Energy is given by

$U=-\frac{GMm}{r}$

Where r is the distance of the body from the surface of the Earth

m is the mass of the body

M is the mass of the Earth

G is gravitational constant $=6.67\times 10^{-11}$ m³ kg⁻¹s⁻²

Here,

$m = 10$ kg

$M =6\times 10^{24}$ kg

$r=6400+6400=12800$ km

Thus,

$U=-\frac{6.67\times 10^{-11}\times 6\times 10^{24}\times 10}{12800\times 10^3}$ Joule

$\implies U=-3.127\times 10^8$ Joule

Hope this answer is helpful.

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