Home
Class 11
PHYSICS
A satellite of mass 50kg orbits the eart...

A satellite of mass 50kg orbits the earth at a height of 100km. Calcualte (i) K.E. (ii) P.E. and (iii) total energy . Given `G = 6.67 xx 10^(-11) Nm^(2) kg^(-2)`. Radius of the arth is 6400 and mass of the earth is `6 xx 10^(24)` kg.

Text Solution

AI Generated Solution

The correct Answer is:
To solve the problem, we need to calculate the kinetic energy (K.E.), potential energy (P.E.), and total energy (T.E.) of a satellite orbiting the Earth. ### Given Data: - Mass of the satellite (m) = 50 kg - Height of the satellite (h) = 100 km = 100,000 m - Radius of the Earth (R) = 6400 km = 6,400,000 m - Gravitational constant (G) = \(6.67 \times 10^{-11} \, \text{Nm}^2/\text{kg}^2\) - Mass of the Earth (M) = \(6 \times 10^{24} \, \text{kg}\) ### Step 1: Calculate the distance from the center of the Earth to the satellite The total distance (r) from the center of the Earth to the satellite is given by: \[ r = R + h = 6,400,000 \, \text{m} + 100,000 \, \text{m} = 6,500,000 \, \text{m} \] ### Step 2: Calculate the Kinetic Energy (K.E.) The formula for the kinetic energy of a satellite in orbit is given by: \[ K.E. = \frac{G \cdot M \cdot m}{2r} \] Substituting the values: \[ K.E. = \frac{(6.67 \times 10^{-11}) \cdot (6 \times 10^{24}) \cdot (50)}{2 \cdot (6,500,000)} \] Calculating the above expression: \[ K.E. = \frac{(6.67 \times 10^{-11}) \cdot (3 \times 10^{26})}{13,000,000} \] \[ K.E. = \frac{2.001 \times 10^{16}}{13,000,000} \approx 15.39 \times 10^{8} \, \text{J} \] ### Step 3: Calculate the Potential Energy (P.E.) The formula for gravitational potential energy is: \[ P.E. = -\frac{G \cdot M \cdot m}{r} \] Substituting the values: \[ P.E. = -\frac{(6.67 \times 10^{-11}) \cdot (6 \times 10^{24}) \cdot (50)}{6,500,000} \] Calculating the above expression: \[ P.E. = -\frac{(6.67 \times 10^{-11}) \cdot (3 \times 10^{26})}{6,500,000} \] \[ P.E. = -\frac{2.001 \times 10^{16}}{6,500,000} \approx -30.77 \times 10^{8} \, \text{J} \] ### Step 4: Calculate the Total Energy (T.E.) The total energy is given by the sum of kinetic and potential energy: \[ T.E. = K.E. + P.E. \] Substituting the values: \[ T.E. = (15.39 \times 10^{8}) + (-30.77 \times 10^{8}) = -15.38 \times 10^{8} \, \text{J} \] ### Final Results: - Kinetic Energy (K.E.) = \(15.39 \times 10^{8} \, \text{J}\) - Potential Energy (P.E.) = \(-30.77 \times 10^{8} \, \text{J}\) - Total Energy (T.E.) = \(-15.38 \times 10^{8} \, \text{J}\)

To solve the problem, we need to calculate the kinetic energy (K.E.), potential energy (P.E.), and total energy (T.E.) of a satellite orbiting the Earth. ### Given Data: - Mass of the satellite (m) = 50 kg - Height of the satellite (h) = 100 km = 100,000 m - Radius of the Earth (R) = 6400 km = 6,400,000 m - Gravitational constant (G) = \(6.67 \times 10^{-11} \, \text{Nm}^2/\text{kg}^2\) - Mass of the Earth (M) = \(6 \times 10^{24} \, \text{kg}\) ...
Promotional Banner

Topper's Solved these Questions

  • GRAVITATION

    ICSE|Exercise FROM THE HUBBLE TELESCOP|2 Videos
  • GRAVITATION

    ICSE|Exercise SELECTED PROBLEMS[FROM ESCAPE VELOCITY ]|20 Videos
  • FRICTION

    ICSE|Exercise Selected problems|30 Videos
  • INTERNAL ENERGY

    ICSE|Exercise SELECTED PROBLEMS (FROM HEAT ENGINES)|21 Videos

Similar Questions

Explore conceptually related problems

What is the work done in taking an object of mass 1kg from the surface of the earth to a height equal to the radius of the earth ? ( G = 6.67 xx 10^(-11) Nm^(2)//Kg^(2) , Radius of the earth = 6400 km, Mass of the earth = 6 xx 10^(24) kg .)

If a spaceship orbits the earth at a height of 500 km from its surface, then determine its (i) kinetic energy, (ii) potential energy, and (iii) total energy (iv) binding energy. Mass of the satellite = 300 kg, Mass of the earth = 6xx10^(24) kg , radius of the earth = 6.4 xx 10^(6) m, G=6.67 xx 10^(-11) "N-m"^(2) kg^(-2) . Will your answer alter if the earth were to shrink suddenly to half its size ?

A satellite orbits the earth at a height of 3.6xx10^(6)m from its surface. Compute it’s a kinetic energy, b. potential energy, c. total energy. Mass of the satellite =500kg mass of the earth =6xx10^(24) kg, radius of the earth =6.4xx10^(6), G=6.67xx10^(-11)Nm^(2)kg^(-2) .

A satellite is revolving round the earth at a height of 600 km. find a. The speed of the satellite and b. The time period of the satellite. Radius of the earth =6400 km and mass of the earth =6xx10^24kg .

If the acceleration due to gravity on earth is 9.81 m//s^(2) and the radius of the earth is 6370 km find the mass of the earth ? (G = 6.67 xx 10^(-11) Nm^(2)//kg^(2))

The radius of the earth is 6.37 xx 10^(6) m and its mean density is 5.5 xx 10^(3) "kg m"^(-3) and G = 6.67 xx 10^(-11) "N-m"^(2) kg^(-2) Find the gravitational potential on the surface of the earth.

Assuming the earth b to be a homogeneous sphere determine the density of the earth from the following data. g = 9.8 m//s^(2), G = 6.67 xx 10^(-11) Nm^(2) kg^(2) , radius of the earth = 6372km.

A satellite revolve round a planet in an orbit just above the surface of a planet. Taking G = 6.67 xx 10^(-11) Nm^(2) kg^(2) and mean density of the planet 5.51 xx 10^(3) kg//m^(3) find the period of the planet.

Suppose there is an artificial satellite revolving round the planet mars at a height of 125 km. What is the orbital velocity of the artificial satellite. Radius of Mars is 3.375 xx 10^(6) m and the mass of the Mars is 6.420 xx 10^(23) kg .

Find the gravitational potential due to a body o fmass 10kg at a distance (i) 10m and (ii) 20 m from the body. Gravitational constant G = 6.67 xx 10^(-11) Nm^(2) kg^(-1) .