If `M_(e)` is the mass of earth and `M_(m)` is the mass of moon `(M_(e)=81 M_(m))`. The potential energy of an object of mass m situated at a distance R from the centre of earth and r from the centre of moon, will be :-

Options

A. `-GmM_(m) (R/81 +r) 1/R^(2)`
B. `-GmM_(e) (81/r+1/R)`
C. `-GmM_(m) (81/R+1/r)`
D. `GmM_(m) (81/R-1/r)`

Answer

To solve the problem, we need to calculate the potential energy of an object of mass \( m \) situated at a distance \( R \) from the center of the Earth and \( r \) from the center of the Moon. We will use the formula for gravitational potential energy, which is given by:

\[ PE = -\frac{G M m}{d} \]

where \( PE \) is the potential energy, \( G \) is the gravitational constant, \( M \) is the mass of the celestial body, \( m \) is the mass of the object, and \( d \) is the distance from the center of the celestial body.

### Step 1: Calculate the potential energy due to Earth
The potential energy due to the Earth is given by:

\[ PE_e = -\frac{G M_e m}{R} \]

where \( M_e \) is the mass of the Earth and \( R \) is the distance from the center of the Earth.

### Step 2: Calculate the potential energy due to Moon
The potential energy due to the Moon is given by:

\[ PE_m = -\frac{G M_m m}{r} \]

where \( M_m \) is the mass of the Moon and \( r \) is the distance from the center of the Moon.

### Step 3: Combine the potential energies
The total potential energy \( PE \) of the object is the sum of the potential energies due to the Earth and the Moon:

\[ PE = PE_e + PE_m \]

Substituting the expressions from Steps 1 and 2:

\[ PE = -\frac{G M_e m}{R} - \frac{G M_m m}{r} \]

### Step 4: Substitute the relation between the masses of Earth and Moon
Given that \( M_e = 81 M_m \), we can substitute this into the equation:

\[ PE = -\frac{G (81 M_m) m}{R} - \frac{G M_m m}{r} \]

### Step 5: Factor out common terms
Factoring out \( -G M_m m \):

\[ PE = -G M_m m \left( \frac{81}{R} + \frac{1}{r} \right) \]

### Final Expression
Thus, the potential energy of the object is:

\[ PE = -G M_m m \left( \frac{81}{R} + \frac{1}{r} \right) \]