Question

If $${}^{\prime }{g}^{\prime }$$ is the acceleration due to gravity on earth then the increase in P.E of a body of mass $${}^{\prime }{m}^{\prime }$$ up to a distance equal to the radius of earth from the earth surface will be?

Answer 1

Hint: The potential energy associated with gravitational force, such as rising items against the Earth's gravity, is known as gravitational energy. The gravitational potential energy of a system of respective masses at a certain distance, using gravitational constant $G$ is $U=-{\displaystyle \frac{GMm}{r}}$ .

Complete answer:
Let us consider;
Radius of the earth is given by $R$
Mass of the earth is given by $M$
Mass of the given body will be $m$ and
Acceleration due to gravity on earth is given by $g$
When a body is on the surface of the earth, we can write the force of attraction of the earth on it.
$U_1=-{\displaystyle \frac{GMm}{R}}......\left(1\right)$
Where, $G=$ Gravitational constant.
When the body is at a height of $$h$$ from the earth's surface, the PE of the system is given by
$U_2=-{\displaystyle \frac{GmM}{R+h}}$
The potential energy of the object at height $h=R$ from the surface of the earth.
$U_2=-{\displaystyle \frac{GmM}{R+R}}=-{\displaystyle \frac{GmM}{2R}}......\left(2\right)$
As a result of the displacement of the body of mass m from the surface to a height equal to the radius $$\left(R\right)$$ of the earth, the rise in PE is given by-
$$\begin{gathered} \mathrm{\Delta }U=U_2-U_1 \\ \mathrm{\Delta }U=-{\displaystyle \frac{GMm}{R+R}}+{\displaystyle \frac{GMm}{R}} \\ \mathrm{\Delta }U=-{\displaystyle \frac{GMm}{2R}}+{\displaystyle \frac{GMm}{R}} \\ \mathrm{\Delta }U={\displaystyle \frac{1}{2}}{\displaystyle \frac{GMm}{R}} \end{gathered}$$
But we know that $GM=g{R}^2$
Hence, $\mathrm{\Delta }U={\displaystyle \frac{1}{2}}{\displaystyle \frac{g{R}^{2}m}{R}}$
Now, one $R$ will be cancelled out so the equation will become
$$\begin{gathered} \mathrm{\Delta }U={\displaystyle \frac{1}{2}}gRm \\ \therefore \mathrm{\Delta }U={\displaystyle \frac{1}{2}}mgR \end{gathered}$$
Therefore the increase in P.E of a body of mass $${}^{\prime }{m}^{\prime }$$ up to a distance equal to the radius of earth from the earth surface will be ${\displaystyle \frac{1}{2}}mgR$.

Note: It should be noted that the formula for gravitational acceleration is $$g={\displaystyle \frac{GM}{r^2}}$$ . It is determined by the earth's mass and radius. This enables us to comprehend the following:-
1. Gravity accelerates all bodies at the same rate, regardless of their mass.
2. It's worth on Earth is determined by the mass of the planet, not the mass of the object.