Advertisements
Advertisements
प्रश्न
Earth revolves around the Sun at 30 km s−1. Calculate the kinetic energy of the Earth. In the previous example, you calculated the potential energy of the Earth. What is the total energy of the Earth in that case? Is the total energy positive? Give reasons.
Advertisements
उत्तर
Given: V = 30 km s−1
Potential energy = − 49.84 × 1032 J
Mass of earth = 5.9 × 1024
Formula:
Kinetic energy (K.E.) = `1/2"MV"^2`
= `1/2 xx 5.9 xx 10^24 xx 30 xx 30 xx 10^32`
= `1/2 xx 5.9 xx 3 xx 3 xx 10^32`
K.E. = 26.55 × 1032 J
∴ T.E. = K.E. – P.E.
T.E. = 26.55 × 1032 – 49.84 × 1032
T.E. = – 23.29 × 1032 J
Total energy is negative. It implies Earth is bounded with Sun.
APPEARS IN
संबंधित प्रश्न
The gravitational potential energy of the Moon with respect to Earth is ____________.
The magnitude of the Sun’s gravitational field as experienced by Earth is
An object of mass 10 kg is hanging on a spring scale which is attached to the roof of a lift. If the lift is in free fall, the reading in the spring scale is ___________.
What is meant by the superposition of the gravitational field?
What is the difference between gravitational potential and gravitational potential energy?
Prove that at points near the surface of the Earth, the gravitational potential energy of the object is U = mgh.
The work done by Sun on Earth at any finite interval of time is
If the Earth’s pull on the Moon suddenly disappears, what will happen to the Moon?
If the ratio of the orbital distance of two planets `"d"_1/"d"_2` = 2, what is the ratio of gravitational field experienced by these two planets?
What is the gravitational potential energy of the Earth and Sun? The Earth to Sun distance is around 150 million km. The mass of the Earth is 5.9 × 1024 kg and the mass of the Sun is 1.9 × 1030 kg.
