Free Gravitation MCQs with Answers

251. A person weighting 400N at the surface of earth at particular location. When the same object is taken to a point located at higher attitude then the weight of the same person will be:

• A. Less than 400 N
• B. Less than 500 N
• C. Equal to 500 N
• D. Zero

Explanation: When the same object is taken to a point located at a higher altitude, the weight of the person will be less than 400 N. This is because weight is determined by the gravitational force exerted by the Earth, and as you move away from the surface, the gravitational force decreases. So, at a higher altitude, the person would weigh less than 400 N.

252. The absolute potential energy of a spaceship on the surface of the earth is:

• A. GmMe/Re^2
• B. GMe/Re
• C. GmMe/2Re
• D. -GMe/Re

Explanation: The absolute potential energy (U) of an object, such as a spaceship, on the surface of the Earth can be calculated using the formula:U = -G * (m * M_E) / r_Ewhere:- G is the gravitational constant- m is the mass of the spaceship- M_E is the mass of the Earth- r_E is the radius of the EarthThis formula represents the gravitational potential energy of the spaceship due to its position on the Earth's surface. The negative sign indicates that the potential energy is lower on the surface than at infinity

253. The orbital speed of satellite orbiting around the Earth is:

• A. √GM/Re
• B. √GMe/Re
• C. √GMe/R²
• D. √GMe/h

Explanation: The correct formula to calculate the orbital speed (v) of a satellite orbiting the Earth is: v = √(GMe/r) where G is the gravitational constant, Me is the mass of the Earth, and r is the distance between the center of the Earth and the satellite. To derive this formula, we can start with the centripetal force equation F = m v² / r G (Me * m) / r² = m v² / r Solving for v, we get: v = √(GMe/r)

254. The time period of communication satellites is:

• A. 1 hour
• B. 2 hour
• C. 12 hours
• D. 24 hours

Explanation: 24 hours: This option is correct. Geostationary communication satellites orbit the Earth at a height of about 35,786 kilometers, taking 24 hours to complete one orbit. This allows them to remain fixed in the sky relative to an observer on Earth, providing continuous communication coverage.

255. The escape velocity from the earth gravitational field depends upon:earth's

• A. rotation of earth
• B. mass of body
• C. radius of earth
• D. mass of earth

Explanation: Escape velocity, Ve =√( 2GM/R) Acceleration due to gravity, g= GM/R², which is always constant. Plugging this value of g in the first equation: Ve= √2gR , where R is the radius of the earth . From the above equation, we see escape velocity is dependent upon is gravitational acceleration and radius of the earth so option C is correct.

256. The Geostationary satellites are

• A. stationary
• B. Rotating with the speed of earth
• C. rotating very fastly
• D. rotating very slowly

Explanation: Geostationary satellites remain at the same position with respect to the Earth, so that they scan the same place in a better way. Therefore, the relative velocity of geostationary satellite with respect to the spinning motion of the Earth is 0 m/s.

257. Satellites revolve around the earth in a circular orbit. What is the relationship between the radius of their orbits and their speeds?

• A. V ∞ r2
• B. V ∞ r
• C. V1 ∞ 1/r
• D. V ∞ 1/r2

Explanation: d) V ∞ 1/r^2 (V is proportional to 1/r^2):This option correctly represents the relationship between the speed (V) of a satellite and the radius (r) of its orbit. The speed of a satellite is inversely proportional to the square of the radius of its orbit. This means that as the radius of the orbit increases, the speed of the satellite decreases, and vice versa.

258. A satellite is orbiting close to the surface of the earth, its speed is

• A. √2gR
• B. √Rg
• C. Rg/2
• D. Rg

Explanation: The force on a mass m, having an angular velocity ω orbiting in a circular path of radius R is F=mRω2. But v=ω R, where v is the linear velocity so F=mv2/R. Equating this with the weight of the body we get, mg=mv2 /R. Or, v2 = Rg. Or, v=√(Rg)

259. Two satellites of masses "3M" and "M" orbit the earth in a circular orbit of radius Y and "3r" respectively, the ratio of their speed is:

• A. 1:1
• B. Square root of 3 :1
• C. 3:1
• D. 9:1

Explanation: To move in a circular orbit, gravitational force provides centripetal force. Gravitational force between the two is the same as it is a mutual force. For the same reason, distance between the satellites is the same so speed does not depend on radius.

260. Once the space shuttle is in orbit at a radius R from earths's centre what force does the seat exert on the astronout?

• A. mg
• B. Zero newton
• C. m/g
• D. mg/R2

Explanation: Zero. The force of gravity is counterbalanced by the force of angular acceleration.That's how it stays in orbit.Have you ever seen any commentary on astronauts in spaceships? They are floating in a spaceship too, with no force.