please give me 8 numericals related to gravitation chapter class 9
Answers
Answer:
Problem 1:
An object is dropped, with no initial velocity, above the surface of planet Big Alpha and falls 13.5 meters in 3 seconds. The radius of planet Big Alpha is 5.82×106 meters.
a) What is the acceleration of the falling object?
b) What is the mass of planet Big Alpha?
Solution to Problem 1:
a) Given the distance and the time, we can calculate the acceleration a using the distance formula for the uniform acceleration motion as follows:
d = (1/2) a t 2
a = 2 d / t 2 = 2 × 13.5 / 3 2 = 3 m/s2
b) Let R be the radius and mb be the mass of planet Big Alpha and mo the mass of the object. The acceleration is due to the universal force of gravity, therefore the universal force of gravity and Newton's second law give
G mb mo / R2 = mo a
Simplify to obtain
mb = a R2 / G = 3 (5.82×106)2 / (6.674×10-11) = 1.52×1024 Kg
Problem 2:
An object is dropped, with no initial velocity, near the surface of planet Manta reaches a speed of 21 meters/seconds in 3.0 seconds. Planet Manta has a mass of 2.3 × 1023 Kg.
a) What is the acceleration acting on the object?
b) What is the radius of planet Manta?
Solution to Problem 2:
a) Given the velocity and the time, we can calculate the acceleration a using the velocity formula of the uniform acceleration motion as follows:
v = a t
a = v / t = 21 / 3 = 7 m/s2
b) Let R be the radius and mm be the mass of planet Manta and mo the mass of the object. The acceleration is due to the universal force of gravity, therefore the universal force of gravity and Newton's second law give
G mm mo / R2 = mo a
Simplify to obtain
R2 = G mm / a
R = √ ( G mm / a ) = √ [ ( 6.674×10-11)(2.3 × 1023) / 7 ] = 1.48 × 106 m
Problem 3:
A 1500 kg satellite orbits the Earth at an altitude of 2.5×106 m.
a) What is the obital speed of the satellite?
b) What is period of the satellite?
c) What is the kinetic energy of the satellite?
Solution to Problem 3:
a) For the satellite to be and stay in orbit, the centripetal Fc and universal Fu forces have to be equal in magnitude.
Fc = m v2 / R , v orbital speed of satellite, m mass of the satellite and R orbital radius
R = Radius of Earth + altitutde = 6.4×106 m + 2.5×106 m = 6.9×106 m
Fu = G M m / R2 , M mass of planet Earth
G M m / R2 = m v2 / R
Simplify to obtain
v = √ (G M / R) = √ [ (6.67×10-11)(5.96×1024)/(6.9×106) ] = 7590 m/s
b) The period T is the time it takes the satellite to complete one rotation around the Earth. Hence
T = 2πR / v = 2π×6.371×106 / 7590 = 5274 s
c) The kinetic energy Ek of the satellite is given by
Ek = (1 / 2) m v2 = (1/2) × 1500 × 75902 = 4.32 × 1010 J
Problem 4:
The force of gravity that acts on an object on the surface of Mars is 20 N. What force of gravity will act on the same object on the surface of the Earth? (use gravitational field strength g = 9.8 N/Kg on the surface of the Earth).
Solution to Problem 4:
Let the gravitational field strength on Mars be gm and that of Earth be g and m be the mass of the object. The acceleration is due to the universal force of gravity, therefore Newton's second law and the universal force of gravity are equal.
m gm = G M m / Rm2 , on the surface of Mars
Simplify to obtain
gm = G M / Rm2
where M (= 6.39 × 1023kg) is the mass of Mars, Rm (= 3.39 × 106m) is radius of Mars.
On the surface of Mars
F = m gm and F = 20 N
m = F / gm = 20 / gm
On the surface of the Earth
Fe = g m = 9.8 × F / gm = 9.8 × 20 / (G M / Rm2) = 9.8×20 × Rm2 / (G M)
= 9.8×20 × (3.39 × 106)2 / (6.674 × 10-11 × 6.39 × 1023) = 53 N
Problem 5:
A 500 Kg satellite was originally placed into an orbit of radius 24,000 km and a period of 31 hours around planet Barigou.
a) Express the mass of this planet in terms of the Universal constant G, the radius R and the period T.
b) The satellite was then put into its final orbit of radius 10,000km. What was its new period?
c) What is the change in the kinetic energy of the satellite from the first to the second orbits?
Problem 6:
A 1000 Kg satellite is in synchronous orbit around planet earth. The period of this synchornous orbit matches the rotation of the earth around its axis, assumed to be 24 hours, so that the satellite appears stationary.
a) What is the orbital radius of the satellite?
b) What is the altitude of the satellite?
c) What is the kinetic of the satellite?
Problem 7:
The gravitational potential energy of a 500 kg satellite, orbiting around a planet of mass 4.2 × 1023, is - 4.8 × 109 J.
a) What is the orbital radius of this satellite?
b) What is the kinetic energy of this satellite?
c) What is the total energy of this satellite?
d) What is orbital speed of this satellite?
Problem 8:
What is the period of a satellite orbiting the moon at an altitude of 5.0 × 103 km.
Problem 9:
What is the acceleration on the surface of the Moon
Problem 10:
The Hubble Space Telescope orbits the Earth at an altitude of 568 km.
a) What is the orbital speed of the telescope?
b) What is the period of the telescope?