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
If the masses of the Earth and Sun suddenly double, the gravitational force between them will ___________.
The gravitational potential energy of the Moon with respect to Earth is ____________.
If the mass and radius of the Earth are both doubled, then the acceleration due to gravity g'
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 ___________.
Will the angular momentum of a planet be conserved? Justify your answer.
Write unit of the gravitational field.
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?
Calculate the gravitational field at point O due to three masses m1, m2 and m3 whose positions are given by the following figure. If the masses m1 and m2 are equal what is the change in a gravitational field at the point O?
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.
