
If \['g'\] is the acceleration due to gravity on earth then the increase in P.E of a body of mass \['m'\] up to a distance equal to the radius of earth from the earth surface will be?
Answer
478.5k+ views
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 $U = - \dfrac{{GMm}}{r}$ .
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.
${U_1} = - \dfrac{{GMm}}{R}......\left( 1 \right)$
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
${U_2} = - \dfrac{{GmM}}{{R + h}}$
The potential energy of the object at height $h = R$ from the surface of the earth.
${U_2} = - \dfrac{{GmM}}{{R + R}} = - \dfrac{{GmM}}{{2R}}......\left( 2 \right)$
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-
$
\Delta U = {U_2} - {U_1} \\
\Delta U = - \dfrac{{GMm}}{{R + R}} + \dfrac{{GMm}}{R} \\
\Delta U = - \dfrac{{GMm}}{{2R}} + \dfrac{{GMm}}{R} \\
\Delta U = \dfrac{1}{2}\dfrac{{GMm}}{R} \\
$
But we know that $GM = g{R^2}$
Hence, $\Delta U = \dfrac{1}{2}\dfrac{{g{R^2}m}}{R}$
Now, one $R$ will be cancelled out so the equation will become
\[
\Delta U = \dfrac{1}{2}gRm \\
\therefore \Delta U = \dfrac{1}{2}mgR \\
\]
Therefore the increase in P.E of a body of mass \['m'\] up to a distance equal to the radius of earth from the earth surface will be $\dfrac{1}{2}mgR$.
Note: It should be noted that the formula for gravitational acceleration is \[g = \dfrac{{GM}}{{{r^2}}}\] . 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.
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.
${U_1} = - \dfrac{{GMm}}{R}......\left( 1 \right)$
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
${U_2} = - \dfrac{{GmM}}{{R + h}}$
The potential energy of the object at height $h = R$ from the surface of the earth.
${U_2} = - \dfrac{{GmM}}{{R + R}} = - \dfrac{{GmM}}{{2R}}......\left( 2 \right)$
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-
$
\Delta U = {U_2} - {U_1} \\
\Delta U = - \dfrac{{GMm}}{{R + R}} + \dfrac{{GMm}}{R} \\
\Delta U = - \dfrac{{GMm}}{{2R}} + \dfrac{{GMm}}{R} \\
\Delta U = \dfrac{1}{2}\dfrac{{GMm}}{R} \\
$
But we know that $GM = g{R^2}$
Hence, $\Delta U = \dfrac{1}{2}\dfrac{{g{R^2}m}}{R}$
Now, one $R$ will be cancelled out so the equation will become
\[
\Delta U = \dfrac{1}{2}gRm \\
\therefore \Delta U = \dfrac{1}{2}mgR \\
\]
Therefore the increase in P.E of a body of mass \['m'\] up to a distance equal to the radius of earth from the earth surface will be $\dfrac{1}{2}mgR$.
Note: It should be noted that the formula for gravitational acceleration is \[g = \dfrac{{GM}}{{{r^2}}}\] . 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.
Recently Updated Pages
Why are manures considered better than fertilizers class 11 biology CBSE

Find the coordinates of the midpoint of the line segment class 11 maths CBSE

Distinguish between static friction limiting friction class 11 physics CBSE

The Chairman of the constituent Assembly was A Jawaharlal class 11 social science CBSE

The first National Commission on Labour NCL submitted class 11 social science CBSE

Number of all subshell of n + l 7 is A 4 B 5 C 6 D class 11 chemistry CBSE

Trending doubts
What is meant by exothermic and endothermic reactions class 11 chemistry CBSE

10 examples of friction in our daily life

One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

1 Quintal is equal to a 110 kg b 10 kg c 100kg d 1000 class 11 physics CBSE

Difference Between Prokaryotic Cells and Eukaryotic Cells

What are Quantum numbers Explain the quantum number class 11 chemistry CBSE

