Question

STANDING
Q13.what will be the P. E of a body of mass 67 kg at a distance of 6.6*10power10 m from
the centre of the earth. Find gravitational potential at this distance.​

Answer 1

Answer:

• The gravitational potential energy at this distance is 40.02 × 10³ J.

Explanation:

Given that,

• Mass of body (m) = 67 kg
• Distance (r) = 6.6 × 10^{10} m
• Mass of earth (M) = 6 × 10^{24} kg
• Gravitational constant (G) = 6.67 × 10^{-11} Nm²kg-²

As we know that,

$\purple\bigstar$Gravitational potential energy = GMm/r

➡ U = GMm/r

[ Putting values ]

➡ U = 6.67 × 10^{-11} × 6 × 10^{-24} × 67/6.6 × 10^{10}

➡ U = 6.67 × 6 × 6.6 × 10^{-11 + 24}/6.6 × 10^{10}

➡ U = 264.132 × 10^{13}/6.6 × 10^{10}

➡ U = 40.02 × 10^{13 - 10}

➡ U = 40.02 × 10³ J $\red\bigstar$

For information:

• K.E = 1/2 mv²
• P.E = mgh
• M.E = P.E + K.E

Answer 2

Answer:

QUESTION

Q13.what will be the P. E of a body of mass 67 kg at a distance of 6.6*10power10 m from

the centre of the earth. Find gravitational potential at this distance.

To find

Gravitational potential at the distance

Solution

➡️Gravitational potential energy = GMm/r

➡️U = GMm/r

➡️ After putting values

➡️U = 6.67 × 10^{-11} × 6 × 10^{-24} × 67/6.6 × 10^{10}

➡️ U = 6.67 × 6 × 6.6 × 10^{-11 + 24}/6.6 × 10^{10}

➡️ U = 264.132 × 10^{13}/6.6 × 10^{10}

➡️ U = 40.02 × 10^{13 - 10}

➡️ U = 40.02 × 10³ J

$\small \fbox {Gravitational potential at the distance = 40.02 × 10³}$