Raising the orbit of the Mars Orbiter Mission

Adapted from IOAA 2016 (Bhubaneswar), theory problem 9. We have restated it in our own words and kept the original data. The solution was last checked on 21 October 2026.

India's Mars Orbiter Mission (MOM) was launched on 5 November 2013. Its dry mass, the body and the instruments, was 500 kg, and it carried 852 kg of fuel. It was first placed on an elliptical orbit round Earth with its perigee 264.1 km and its apogee 23 903.6 km above the surface. After six raisings of the orbit it was sent on its transfer orbit to Mars.

The first raising was made by firing the engines for a very short time near perigee, without changing the plane of the orbit or its perigee. This gave the spacecraft an impulse of $1.73\times {10}^5$ kg m s⁻¹. Ignore the loss of mass as the fuel burned. Take Earth's radius as 6371 km, its mass as $5.972\times {10}^{24}$ kg and $G=6.674\times {10}^{-11}$ N m² kg⁻².

(a) What is the height $h_{\text{a}}$ of the new apogee above Earth's surface? [14 points]

(b) Find the eccentricity of the new orbit and its period in hours. [6 points]

Answer(a) The new apogee is about 28 860 km above the surface. (b) The eccentricity is 0.683 and the period 8.37 hours. The official solution gives 29 009 km, about 150 km higher, because it rounds the speeds and the ratio along the way. The apogee depends on the small difference 0.1883, so every rounding is magnified, and we keep four figures until the end.